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Biomolecules - Monosaccharides (Glucose and Fructose)

Grade 12ICSEChemistry

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

🔑Concepts

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

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

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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.

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

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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).

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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.

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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.

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

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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+H2O→H+C6H12O6⏟Glucose+C6H12O6⏟FructoseC_{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+nH2O→393K,2−3atmH+nC6H12O6(C_6H_{10}O_5)_n + nH_2O \xrightarrow[393K, 2-3 atm]{H^+} nC_6H_{12}O_6

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

CHO−(CHOH)4−CH2OH+Br2/H2O→COOH−(CHOH)4−CH2OH (Gluconic acid)CHO-(CHOH)_4-CH_2OH + Br_2/H_2O \rightarrow COOH-(CHOH)_4-CH_2OH \text{ (Gluconic acid)}

CHO−(CHOH)4−CH2OH→HNO3COOH−(CHOH)4−COOH (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: C6H12O6→HI/PCH3−CH2−CH2−CH2−CH2−CH3C_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.