Exhaustive Guide: Carbohydrates | Biomolecules | Class 12 Chemistry | ChemCA

Carbohydrates: Structure & Classification

Module 1 | CBSE Class 12 Chemistry | Biomolecules

1. Introduction to Carbohydrates

Biomolecules are complex, lifeless organic molecules that combine in a specific manner to produce life. Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds.

Modern Definition of Carbohydrates: Carbohydrates are defined as optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Some carbohydrates, which are sweet in taste, are also called sugars.

The most common sugar used in our homes is sucrose, whereas the sugar present in milk is known as lactose.

2. Classification of Carbohydrates

Carbohydrates are classified on the basis of their behavior on hydrolysis:

Monosaccharides: Carbohydrates that cannot be hydrolyzed further to give simpler units of polyhydroxy aldehyde or ketone. Examples: Glucose, Fructose, Ribose.
Oligosaccharides: Yield 2 to 10 monosaccharide units on hydrolysis. They are further classified as disaccharides (e.g., sucrose, maltose), trisaccharides, etc.
Polysaccharides: Yield a large number of monosaccharide units on hydrolysis. They are generally not sweet in taste and are called non-sugars. Examples: Starch, Cellulose, Glycogen.

2.1 Reducing vs. Non-Reducing Sugars (Board Favorite)

Reducing Sugars: Carbohydrates that can reduce Tollens' reagent and Fehling's solution. This happens because they contain a free aldehyde or ketone group.
Examples: All monosaccharides (both aldoses and ketoses), Maltose, Lactose.

Non-Reducing Sugars: Carbohydrates in which the reducing groups (aldehyde or ketone) are bonded together in the glycosidic linkage, leaving no free functional group to act as a reducing agent.
Example: Sucrose.

3. Monosaccharides: Glucose (Aldohexose)

Glucose is an aldohexose. It is the monomer of many of the larger carbohydrates, namely starch, cellulose. It is also known as dextrose. The molecular formula of glucose was found to be C₆H₁₂O₆.

3.1 Structure Elucidation of D-Glucose (Crucial for Boards)

The open-chain structure of glucose was established based on the following classical chemical reactions. (You must memorize these reagents and what they prove):

Prolonged heating with HI:
Yields n-hexane.
Conclusion: All six carbon atoms are linked in a straight chain.
Reaction with Hydroxylamine (NH₂OH) & HCN:
Forms an oxime with NH₂OH and a cyanohydrin with HCN.
Conclusion: Confirms the presence of a carbonyl group (>C=O).
Reaction with Bromine water (mild oxidizing agent):
Glucose gets oxidized to a six-carbon carboxylic acid called gluconic acid.
Conclusion: Confirms that the carbonyl group is an aldehyde (-CHO) group.
Reaction with Acetic anhydride:
Yields glucose pentaacetate.
Conclusion: Confirms the presence of five -OH groups. Since it is stable, the five -OH groups must be attached to different carbon atoms.
Reaction with Nitric acid (HNO₃, strong oxidizing agent):
Both glucose and gluconic acid yield a dicarboxylic acid called saccharic acid.
Conclusion: Confirms the presence of a primary alcoholic group (-CH₂OH) at the terminal end.

Note on D/L Nomenclature: The capital 'D' in D-(+)-glucose signifies the spatial configuration. It means the -OH group on the lowest chiral carbon (C-5) is on the right side, homologous to D-glyceraldehyde. The '(+)' represents dextrorotatory optical activity.

3.2 Cyclic Structure & Anomers

Despite the open-chain structure explaining many properties, it failed to explain several anomalies:

Glucose does not give the 2,4-DNP test, Schiff’s test, and it does not form the hydrogensulphite addition product with NaHSO₃.
The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free -CHO group.
Glucose exists in two different crystalline forms: α and β.

The Cyclic Hemiacetal: To explain these, it was proposed that one of the -OH groups (specifically at C-5) adds to the -CHO group, forming a cyclic hemiacetal structure. This consumes the free aldehyde group.

Anomers: The two cyclic hemiacetal forms of glucose (α and β) differ only in the configuration of the hydroxyl group at C-1. This C-1 carbon is called the anomeric carbon. Such isomers, which differ in configuration only at the anomeric carbon, are called anomers.

- α-D-Glucose: The -OH group at C-1 is towards the right (or pointing down in Haworth projection).
- β-D-Glucose: The -OH group at C-1 is towards the left (or pointing up in Haworth projection).

The six-membered cyclic structure of glucose is called a Pyranose structure (analogous to the organic compound pyran).

Fructose (Ketohexose): Fructose also exists in two cyclic forms, but it forms a five-membered ring called a Furanose ring by the addition of the -OH at C-5 to the ketone group at C-2. Therefore, in fructose, C-2 is the anomeric carbon.

4. Disaccharides & Glycosidic Linkages

Disaccharides yield two molecules of the same or different monosaccharides on hydrolysis with dilute acids or enzymes.

Glycosidic Linkage: The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through an oxygen atom is called a glycosidic linkage.

4.1 Sucrose (Invert Sugar)

Sucrose (C₁₂H₂₂O₁₁) yields D-glucose and D-fructose upon hydrolysis.

Linkage: The glycosidic linkage is between C-1 of α-D-glucose and C-2 of β-D-fructose.
Reducing Property: Since the reducing groups of both glucose (C-1) and fructose (C-2) are involved in glycosidic bond formation, sucrose is a non-reducing sugar.

Concept of Invert Sugar:
Sucrose itself is dextrorotatory (+66.5°). However, upon hydrolysis, it produces dextrorotatory D-glucose (+52.5°) and strongly laevorotatory D-fructose (-92.4°). Because the laevorotation of fructose exceeds the dextrorotation of glucose, the overall mixture becomes laevorotatory.
Thus, hydrolysis of sucrose brings about a change in the sign of rotation (from dextro to laevo). The resulting mixture is called invert sugar.

4.2 Maltose & Lactose

Maltose: Composed of two α-D-glucose units. The linkage is between C-1 of one glucose and C-4 of another. The free aldehyde group can be produced at C-1 of the second glucose in solution, making it a reducing sugar.
Lactose (Milk Sugar): Composed of β-D-galactose and β-D-glucose. The linkage is between C-1 of galactose and C-4 of glucose. It is a reducing sugar.

5. Polysaccharides (Starch, Cellulose, Glycogen)

Polysaccharides contain a large number of monosaccharide units joined by glycosidic linkages. They act as food storage or structural materials.

1. Starch: Main storage polysaccharide of plants. It is a polymer of α-glucose and consists of two components:
- Amylose (15-20%): Water soluble. A long unbranched chain of 200-1000 α-D-glucose units held by C1-C4 α-glycosidic linkages.
- Amylopectin (80-85%): Water insoluble. A branched chain polymer. Linear chain is formed by C1-C4 glycosidic linkages, whereas branching occurs via C1-C6 glycosidic linkages.
2. Cellulose: Predominant constituent of the cell wall of plant cells. It is a straight-chain polysaccharide composed exclusively of β-D-glucose units joined by glycosidic linkages between C-1 of one glucose and C-4 of the next.
3. Glycogen (Animal Starch): The carbohydrates are stored in the animal body as glycogen. Its structure is highly branched, very similar to amylopectin, but more extensively branched.

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 14.1: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.

Solution:
Glucose contains five hydroxyl (-OH) groups, and sucrose contains eight -OH groups. These highly polar -OH groups undergo extensive intermolecular hydrogen bonding with water molecules, making them highly soluble in water.
Cyclohexane and benzene are non-polar hydrocarbons. They lack -OH groups and cannot form hydrogen bonds with water, making them insoluble in polar solvents like water.

NCERT Example 14.2: What are the expected products of hydrolysis of lactose?

Solution:
Lactose is a disaccharide. Upon acidic or enzymatic hydrolysis, it breaks the C1-C4 glycosidic linkage to yield equimolar amounts of its constituent monosaccharides: β-D-galactose and β-D-glucose.

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2021]

Q1. What happens when D-glucose is treated with the following reagents?
(i) HI
(ii) Bromine water
(iii) HNO₃

Answer:
(i) Prolonged heating with HI forms n-hexane, indicating that all 6 carbon atoms are in a straight chain.
(ii) Bromine water (a mild oxidizing agent) oxidizes the terminal aldehyde group to a carboxyl group, forming gluconic acid.
(iii) Nitric acid (a strong oxidizing agent) oxidizes both the terminal aldehyde and the terminal primary alcohol (-CH₂OH) to carboxyl groups, forming saccharic acid.

[CBSE 2017, 2020]

Q2. Why is sucrose called a non-reducing sugar, while maltose is a reducing sugar?

Answer: A sugar is reducing if it has a free aldehyde or ketone group (hemiacetal/hemiketal form). In sucrose, the glycosidic linkage is formed between the reducing C-1 of α-glucose and the reducing C-2 of β-fructose. Since both reducing groups are tied up in the bond, it cannot reduce Tollens' or Fehling's reagents. In maltose, the linkage is C1-C4. The C-1 of the second glucose unit remains free (hemiacetal), allowing it to act as a reducing sugar.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2024]

Q3. Assertion (A): D-Glucose does not give 2,4-DNP test and Schiff’s test.
Reason (R): D-Glucose exists predominantly in a cyclic hemiacetal structure.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. The 2,4-DNP and Schiff's tests require a free aldehyde group. Because glucose undergoes intramolecular reaction to form a stable cyclic hemiacetal, the free -CHO group is no longer present in sufficient concentration to react with these reagents.

Part C: Structures & Differences (3 Marks)

[CBSE 2016, 2019]

Q4. Explain the following terms:
(a) Invert sugar
(b) Anomers
(c) Structural difference between amylose and amylopectin

Answer:
(a) Invert Sugar: When dextrorotatory sucrose is hydrolyzed, it yields a mixture of D-glucose (+52.5°) and D-fructose (-92.4°). The specific rotation of the mixture becomes laevorotatory. This change in the sign of rotation is called inversion, and the resulting mixture is invert sugar.

(b) Anomers: Cyclic stereoisomers of a monosaccharide (like α and β glucose) that differ in configuration only at the hemiacetal or hemiketal carbon (C-1 in glucose, C-2 in fructose), which is called the anomeric carbon.

(c) Amylose vs Amylopectin: Amylose is a water-soluble, straight-chain polymer of α-D-glucose linked entirely by C1-C4 glycosidic bonds. Amylopectin is a water-insoluble, highly branched polymer where the linear chains are linked by C1-C4 bonds, and the branching occurs via C1-C6 glycosidic linkages.

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Exhaustive Guide: Carbohydrates | Biomolecules | Class 12 Chemistry