Example 1
Derive the formula for a disaccharide formed from two C6H12O6 molecules.
2C6H12O6−H2O=C12H22O11. A single glycosidic link forms with loss of one water molecule.
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Year 12 · SACE Stage 2
Carbohydrate structures and condensation
Recognise carbohydrate features and calculate linked-sugar formulae.
Method and worked examples
Carbohydrates include polyhydroxy aldehydes, polyhydroxy ketones and substances producing these units on hydrolysis. Glucose has molecular formula C6H12O6 and exists mainly in cyclic forms in water, in equilibrium with a small open-chain fraction. The open-chain aldehyde form helps explain its reducing behaviour. A simple carbon–hydrogen–oxygen ratio alone is not a sufficient definition of a carbohydrate.
Condensation joins sugar units through glycosidic links while releasing water. Joining two glucose molecules into a disaccharide removes one H2O, giving C12H22O11. A long glucose-based polysaccharide is commonly represented by repeating unit (C6H10O5)n; exact finite chains include end groups. Hydrolysis reverses link formation. Starch and glycogen store energy, while cellulose provides structural support; differing link arrangements give different properties despite related monomers.
Example 1
Example 2
How many water molecules are released to make a single unbranched chain of five glucose units?
Five monomers require four connecting links, so four H2O molecules are released. In general a finite chain of n units has n−1 links.
Watch out: Many fit a related ratio, but the class is defined by structure and hydrolysis products; deoxyribose is a counterexample to that simple formula.