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Biomolecules - Monosaccharides (Glucose and Fructose)

Grade 12ICSEChemistry

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

🔑Concepts

Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed into smaller units. Their general formula is (CH2O)n(CH_2O)_n.

Glucose (C6H12O6C_6H_{12}O_6) is an aldohexose, meaning it contains an aldehyde group (CHO-CHO) and six carbon atoms.

The open-chain structure of Glucose was established through chemical reactions: heating with HIHI yields nn-hexane, proving a straight chain of six carbons; reaction with NH2OHNH_2OH yields an oxime, proving a carbonyl group.

Mild oxidation of Glucose with bromine water (Br2/H2OBr_2/H_2O) produces Gluconic acid, confirming the carbonyl group is an aldehyde.

Strong oxidation of Glucose with nitric acid (HNO3HNO_3) produces Saccharic acid (a dicarboxylic acid), indicating the presence of a primary alcoholic group (CH2OH-CH_2OH).

Glucose exists in two cyclic hemiacetal forms, α\alpha-DD-glucose and β\beta-DD-glucose, which differ only in the configuration of the hydroxyl group at C1C_1. These isomers are called anomers.

Fructose (C6H12O6C_6H_{12}O_6) is a ketohexose. It contains a ketonic functional group at C2C_2 and is naturally laevorotatory, hence often called laevulose.

While Glucose forms a six-membered pyranose ring, Fructose typically forms a five-membered furanose ring in its cyclic state.

Mutarotation is the spontaneous change in the specific rotation of an optically active compound (like α\alpha or β\beta glucose) when dissolved in water, until an equilibrium is reached.

📐Formulae

C12H22O11+H2OH+C6H12O6Glucose+C6H12O6FructoseC_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} \underbrace{C_6H_{12}O_6}_{Glucose} + \underbrace{C_6H_{12}O_6}_{Fructose}

(C6H10O5)n+nH2O393K,23atmH+nC6H12O6(C_6H_{10}O_5)_n + nH_2O \xrightarrow[393K, 2-3 atm]{H^+} nC_6H_{12}O_6

CHO(CHOH)4CH2OH+5(CH3CO)2OGlucose pentaacetate+5CH3COOHCHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow \text{Glucose pentaacetate} + 5CH_3COOH

CHO(CHOH)4CH2OH+Br2/H2OCOOH(CHOH)4CH2OH (Gluconic acid)CHO-(CHOH)_4-CH_2OH + Br_2/H_2O \rightarrow COOH-(CHOH)_4-CH_2OH \text{ (Gluconic acid)}

CHO(CHOH)4CH2OHHNO3COOH(CHOH)4COOH (Saccharic acid)CHO-(CHOH)_4-CH_2OH \xrightarrow{HNO_3} COOH-(CHOH)_4-COOH \text{ (Saccharic acid)}

💡Examples

Problem 1:

What happens when DD-Glucose is treated with HIHI and red phosphorus?

Solution:

nn-hexane is formed: C6H12O6HI/PCH3CH2CH2CH2CH2CH3C_6H_{12}O_6 \xrightarrow{HI/P} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3

Explanation:

This reaction is used to prove that all six carbon atoms in glucose are linked in a straight chain.

Problem 2:

Explain why glucose does not give Schiff's test despite having an aldehyde group.

Solution:

Glucose exists mainly in a cyclic hemiacetal form.

Explanation:

In the cyclic structure, the CHO-CHO group is involved in ring formation (at C1C_1). Only a very small amount of open-chain form exists in equilibrium, which is insufficient to react with Schiff's reagent or form a sodium bisulphite addition product.

Problem 3:

Define Anomers with respect to Glucose.

Solution:

α\alpha-DD-Glucose and β\beta-DD-Glucose are anomers.

Explanation:

Anomers are a specific type of diastereomer that differ in configuration only at the hemiacetal or hemiketal carbon, which is C1C_1 for Glucose and C2C_2 for Fructose. This carbon is known as the anomeric carbon.

Monosaccharides (Glucose and Fructose) Revision - Class 12 Chemistry ICSE