CUET UG Chemistry Booster Test 3 Carbohydrates

Statement 1 is correct because living organisms are made up of non-living atoms and molecules arranged in a highly organized manner. Statement 2 is incorrect because life processes depend on chemical interactions among biomolecules, not isolated non-chemical interactions. Statement 3 is correct because carbohydrates and proteins are major biomolecules that help sustain life through energy supply and structural functions. Therefore, Statements 1 and 3 are correct.

Living organisms remain alive because of the coordinated and synchronized chemical reactions occurring among biomolecules such as carbohydrates, proteins, nucleic acids, and lipids. These reactions maintain metabolism, growth, repair, and reproduction. Therefore, life emerges from the harmonious functioning of biochemical processes.

When glucose reacts with acetic anhydride, all available hydroxyl groups are acetylated. The formation of glucose pentaacetate demonstrates that glucose possesses five distinct hydroxyl groups capable of undergoing acetylation. Therefore, the formation of pentaacetate provides direct evidence for five hydroxyl groups.

Hydroxylamine reacts with aldehydes and ketones to form oximes. Hydrogen cyanide adds across carbonyl groups to form cyanohydrins. Since glucose undergoes both reactions, it must contain a carbonyl group in its open-chain form. Therefore, these reactions confirm the presence of a carbonyl group.

Naturally occurring glucose belongs to the D-series based on its configuration relative to glyceraldehyde. It rotates plane-polarized light to the right, represented by the positive sign (+). Thus, the complete notation is D-(+)-Glucose.

The prefix \\\"tetra\\\" indicates four. Therefore, a tetrasaccharide consists of four monosaccharide units joined by glycosidic linkages. Complete hydrolysis breaks these linkages and yields four monosaccharides.

The passage clearly states that polysaccharides mainly function as food storage materials and structural materials. Examples include starch for storage and cellulose for structural support. Therefore, Option B correctly describes their biological roles.

Starch contains approximately 15–20% amylose and 80–85% amylopectin. Amylopectin is a branched polymer of alpha-D-glucose and is water-insoluble. Therefore, the remaining major component of starch is amylopectin.

Sucrose contains glucose and fructose linked through their reducing groups. Because both the aldehydic and ketonic centers participate in glycosidic bond formation, no free reducing group remains available to react with Tollens' reagent. Hence sucrose behaves as a non-reducing sugar.

Statement 1 is correct because maltose retains a free anomeric carbon capable of generating an aldehyde group. Statement 2 is correct because lactose also retains a free anomeric carbon on the glucose unit. Statement 3 is incorrect because both maltose and lactose are reducing sugars. Therefore, Statements 1 and 2 are correct. When glucose undergoes cyclisation, the aldehydic carbon atom (C1) becomes a new stereogenic centre. The alpha and beta forms of glucose differ only in the orientation of the hydroxyl group attached to this carbon. This special carbon atom is known as the anomeric carbon because it gives rise to the two anomeric forms (α and β).

When glucose undergoes cyclisation, the aldehydic carbon atom (C1) becomes a new stereogenic centre. The alpha and beta forms of glucose differ only in the orientation of the hydroxyl group attached to this carbon.

This special carbon atom is known as the anomeric carbon because it gives rise to the two anomeric forms (α and β).

Although glucose contains an aldehyde group in its open-chain structure, it predominantly exists as a cyclic hemiacetal in solution. The C-5 hydroxyl group reacts intramolecularly with the aldehyde group at C-1 to form a stable six-membered ring. As a result, the free aldehyde concentration is extremely low, causing glucose to fail Schiff\\\'s test and NaHSO₃ addition reactions.

In carbohydrate stereochemistry: • (+)-Glyceraldehyde serves as the standard for D-configuration. • (−)-Glyceraldehyde serves as the standard for L-configuration. • The configuration of the lowest asymmetric carbon atom is compared with glyceraldehyde to assign D or L notation. Thus: 1 → b 2 → a 3 → c

The sugar component of RNA is D-ribose, whereas DNA contains 2-deoxy-D-ribose. These two pentoses form the backbone of nucleic acids and differ by the absence of one oxygen atom at C-2 in deoxyribose.

Oxidation of glucose with nitric acid converts the aldehyde group at C-1 and the primary alcohol group at C-6 into carboxylic acid groups. The resulting dicarboxylic acid is known as saccharic acid. This reaction provided important structural evidence for glucose.

D-ribose is a pentose containing five carbon atoms, whereas D-glucose is a hexose containing six carbon atoms. Therefore, the increasing order of carbon number is: D-ribose < D-glucose

The α-form of glucose is obtained from concentrated solutions at 303 K, whereas the β-form crystallises from hot saturated solutions around 371 K. In aqueous solution, both cyclic forms continuously interconvert through the open-chain form, establishing equilibrium. Hence all three statements are correct.

Cyclisation of glucose produces a six-membered ring containing five carbon atoms and one oxygen atom. This ring closely resembles the heterocyclic compound pyran and is therefore called a pyranose ring.

Starch consists of two components: amylose and amylopectin. Amylose contributes approximately 15–20%, while amylopectin contributes about 80–85% and exists as a branched, water-insoluble polymer.

Statement 1 is correct because cellulose is the major structural component of plant cell walls. Statement 2 is correct because cellulose is extensively used in textile, paper, and lacquer industries. Statement 3 is incorrect because glycogen functions as a storage polysaccharide in animals and does not form plant cell walls. Therefore, Statements 1 and 2 are correct.