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Biomolecules - Carbohydrates (Classification and Structure)

Grade 12CBSEChemistry

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

🔑Concepts

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.

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).

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.

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).

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

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

Sucrose consists of α\alpha-D-glucose and β\beta-D-fructose linked by C1C2C_1-C_2 glycosidic bond. It is a non-reducing sugar and undergoes 'inversion' during hydrolysis.

Starch is a polymer of α\alpha-D-glucose consisting of two components: Amylose (water-soluble, linear, C1C4C_1-C_4 linkage) and Amylopectin (water-insoluble, branched, C1C4C_1-C_4 and C1C6C_1-C_6 linkages).

📐Formulae

Cn(H2O)nC_n(H_2O)_n

C6H12O6+2[Ag(NH3)2]++3OHC5H11O5COO+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)+H2OH+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)

CHO(CHOH)4CH2OHBr2/H2OCOOH(CHOH)4CH2OHCHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2/H_2O} COOH-(CHOH)_4-CH_2OH

CHO(CHOH)4CH2OHHNO3COOH(CHOH)4COOHCHO-(CHOH)_4-CH_2OH \xrightarrow{HNO_3} COOH-(CHOH)_4-COOH

💡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 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 α\alpha-D-glucose with α(14)\alpha(1\rightarrow4) glycosidic linkages and is water-soluble. Amylopectin is a branched polymer with α(14)\alpha(1\rightarrow4) linkages in the chain and α(16)\alpha(1\rightarrow6) linkages at the branching points; it is water-insoluble.

Explanation:

Starch is composed of 1520%15-20\% Amylose and 8085%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 α\alpha-D-glucopyranose and β\beta-D-glucopyranose.

Explanation:

The OH-OH group at C1C_1 is on the right in the α\alpha-form and on the left in the β\beta-form (Fischer projection).