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 .
Classification based on hydrolysis: Monosaccharides (cannot be hydrolyzed further, e.g., ), 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 () is an aldohexose. Its structure was elucidated by reactions: reduction to -hexane (proves straight chain), reaction (proves carbonyl group), and water oxidation to Gluconic acid (proves aldehyde group).
Cyclic structure of Glucose: Glucose exists in two crystalline forms, and , which are anomers. These are cyclic hemiacetals formed by the reaction of the at with the at .
Glycosidic Linkage: The oxide linkage that joins two monosaccharide units through an oxygen atom, involving the loss of a water molecule ().
Sucrose consists of -D-glucose and -D-fructose linked by glycosidic bond. It is a non-reducing sugar and undergoes 'inversion' during hydrolysis.
Starch is a polymer of -D-glucose consisting of two components: Amylose (water-soluble, linear, linkage) and Amylopectin (water-insoluble, branched, and linkages).
📐Formulae
💡Examples
Problem 1:
Explain why glucose reacts with to form an oxime but does not react with .
Solution:
Glucose reacts with because the open-chain form (which contains the free group) exists in equilibrium with the cyclic forms. However, the concentration of the open-chain form is too low to react with , 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 -D-glucose with glycosidic linkages and is water-soluble. Amylopectin is a branched polymer with linkages in the chain and linkages at the branching points; it is water-insoluble.
Explanation:
Starch is composed of Amylose and 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 ( in aldoses). For glucose, these are -D-glucopyranose and -D-glucopyranose.
Explanation:
The group at is on the right in the -form and on the left in the -form (Fischer projection).