Review the key concepts, formulae, and examples before starting your quiz.
πConcepts
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).
πFormulae
π‘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).