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Biomolecules - Carbohydrates: Classification and functions

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Carbohydrates are defined as optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis. They are also known as saccharides.

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Classification based on Hydrolysis: 1. Monosaccharides (cannot be hydrolyzed further, e.g., C6H12O6C_6H_{12}O_6). 2. Oligosaccharides (yield 2-10 units, e.g., Sucrose). 3. Polysaccharides (yield a large number of units, e.g., Starch).

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Classification based on Reducing Nature: Reducing sugars possess free aldehydic or ketonic groups and reduce Tollen's reagent or Fehling's solution (e.g., Glucose, Maltose). Non-reducing sugars do not (e.g., Sucrose).

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Glucose (C6H12O6C_6H_{12}O_6) is an aldohexose. Its structure contains one βˆ’CHO-CHO group, one primary alcoholic group (βˆ’CH2OH-CH_2OH), and four secondary alcoholic groups (βˆ’CHOH-CHOH).

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Anomers are a pair of cyclic stereoisomers (diastereomers) of a sugar that differ only in their configuration at the hemiacetal/hemiketal carbon, known as the anomeric carbon (e.g., Ξ±\alpha-D-glucose and Ξ²\beta-D-glucose).

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Glycosidic Linkage: The oxide linkage formed between two monosaccharide units through the loss of a water molecule (H2OH_2O).

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Inversion of Sugar: Sucrose is dextrorotatory (+66.5∘+66.5^\circ), but after hydrolysis, it gives a mixture of Glucose (+52.5∘+52.5^\circ) and Fructose (βˆ’92.4∘-92.4^\circ), making the final mixture laevorotatory. This is called 'Invert Sugar'.

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Functions: Carbohydrates serve as bio-fuels (Energy source), storage molecules (Starch in plants, Glycogen in animals), and structural material (Cellulose in plant cell walls).

πŸ“Formulae

Cx(H2O)yC_x(H_2O)_y

C12H22O11+H2O→H+C6H12O6(Glucose)+C6H12O6(Fructose)C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 (Glucose) + C_6H_{12}O_6 (Fructose)

C6H12O6β†’HI,Ξ”CH3βˆ’CH2βˆ’CH2βˆ’CH2βˆ’CH2βˆ’CH3(nβˆ’Hexane)C_6H_{12}O_6 \xrightarrow{HI, \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3 (n-Hexane)

[C6H10O5]n+nH2O→nC6H12O6[C_6H_{10}O_5]_n + nH_2O \rightarrow nC_6H_{12}O_6

πŸ’‘Examples

Problem 1:

Show the reaction of Glucose with Hydroxylamine (NH2OHNH_2OH). What does this reaction indicate?

Solution:

C6H12O6+NH2OHβ†’C5H11O5βˆ’CH=Nβˆ’OHC_6H_{12}O_6 + NH_2OH \rightarrow C_5H_{11}O_5-CH=N-OH (Glucose oxime) + H2OH_2O

Explanation:

The formation of an oxime confirms the presence of a carbonyl group (>C=O>C=O) in the glucose molecule.

Problem 2:

Identify the monosaccharide units present in Maltose and the type of linkage connecting them.

Solution:

Maltose is composed of two Ξ±\alpha-D-glucose units linked by an Ξ±(1β†’4)\alpha(1 \rightarrow 4) glycosidic linkage.

Explanation:

The C1C_1 of one glucose unit is linked to C4C_4 of another glucose unit. Since the C1C_1 of the second unit is free, it is a reducing sugar.

Problem 3:

Why is Cellulose not digestible by humans?

Solution:

Human beings lack the enzyme cellulase required to break the Ξ²(1β†’4)\beta(1 \rightarrow 4) glycosidic linkages present in Cellulose.

Explanation:

Cellulose is a straight-chain polysaccharide of Ξ²\beta-D-glucose. Only ruminants (like cows) have bacteria in their digestive tract that secrete cellulase.

Carbohydrates: Classification and functions Class 12 Notes & Examples