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Biomolecules - Carbohydrates (Classification and Structure)

Grade 12CBSEChemistry

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. Their general formula is often represented as Cx(H2O)yC_x(H_2O)_y.

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

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Reducing sugars contain free aldehydic or ketonic groups and can reduce Fehling's solution and Tollen's reagent. Examples include Glucose, Fructose, and Maltose. Non-reducing sugars like Sucrose lack these free groups.

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Glucose (C6H12O6C_6H_{12}O_6) is an aldohexose. Its structure was elucidated by reactions: HIHI reduction to nn-hexane (proves straight chain), NH2OHNH_2OH reaction (proves carbonyl group), and Br2Br_2 water oxidation to Gluconic acid (proves aldehyde group).

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Cyclic structure of Glucose: Glucose exists in two crystalline forms, Ξ±\alpha and Ξ²\beta, which are anomers. These are cyclic hemiacetals formed by the reaction of the βˆ’OH-OH at C5C_5 with the βˆ’CHO-CHO at C1C_1.

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Glycosidic Linkage: The oxide linkage that joins two monosaccharide units through an oxygen atom, involving the loss of a water molecule (H2OH_2O).

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Sucrose consists of Ξ±\alpha-D-glucose and Ξ²\beta-D-fructose linked by C1βˆ’C2C_1-C_2 glycosidic bond. It is a non-reducing sugar and undergoes 'inversion' during hydrolysis.

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Starch is a polymer of Ξ±\alpha-D-glucose consisting of two components: Amylose (water-soluble, linear, C1βˆ’C4C_1-C_4 linkage) and Amylopectin (water-insoluble, branched, C1βˆ’C4C_1-C_4 and C1βˆ’C6C_1-C_6 linkages).

πŸ“Formulae

Cn(H2O)nC_n(H_2O)_n

C6H12O6+2[Ag(NH3)2]++3OHβˆ’β†’C5H11O5COOβˆ’+2Ag+4NH3+2H2OC_6H_{12}O_6 + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_5H_{11}O_5COO^- + 2Ag + 4NH_3 + 2H_2O

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

CHOβˆ’(CHOH)4βˆ’CH2OHβ†’Br2/H2OCOOHβˆ’(CHOH)4βˆ’CH2OHCHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2/H_2O} COOH-(CHOH)_4-CH_2OH

CHOβˆ’(CHOH)4βˆ’CH2OHβ†’HNO3COOHβˆ’(CHOH)4βˆ’COOHCHO-(CHOH)_4-CH_2OH \xrightarrow{HNO_3} COOH-(CHOH)_4-COOH

πŸ’‘Examples

Problem 1:

Explain why glucose reacts with NH2OHNH_2OH to form an oxime but does not react with NaHSO3NaHSO_3.

Solution:

Glucose reacts with NH2OHNH_2OH because the open-chain form (which contains the free βˆ’CHO-CHO group) exists in equilibrium with the cyclic forms. However, the concentration of the open-chain form is too low to react with NaHSO3NaHSO_3, and the cyclic hemiacetal structure is stable enough to resist this specific addition.

Explanation:

This behavior suggests that the aldehyde group in glucose is not completely 'free' and exists primarily in the cyclic hemiacetal form.

Problem 2:

What is the difference between Amylose and Amylopectin?

Solution:

Amylose is a linear polymer of Ξ±\alpha-D-glucose with Ξ±(1β†’4)\alpha(1\rightarrow4) glycosidic linkages and is water-soluble. Amylopectin is a branched polymer with Ξ±(1β†’4)\alpha(1\rightarrow4) linkages in the chain and Ξ±(1β†’6)\alpha(1\rightarrow6) linkages at the branching points; it is water-insoluble.

Explanation:

Starch is composed of 15βˆ’20%15-20\% Amylose and 80βˆ’85%80-85\% Amylopectin.

Problem 3:

Define Anomers with respect to Glucose.

Solution:

Anomers are a pair of optical isomers that differ in configuration only at the anomeric carbon (C1C_1 in aldoses). For glucose, these are Ξ±\alpha-D-glucopyranose and Ξ²\beta-D-glucopyranose.

Explanation:

The βˆ’OH-OH group at C1C_1 is on the right in the Ξ±\alpha-form and on the left in the Ξ²\beta-form (Fischer projection).

Carbohydrates (Classification and Structure) Revision - Class 12 Chemistry CBSE