Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following statements: (1) monosaccharides are optically active polyhydroxy carbonyl compounds (2) fructose does not reduce Fehling's solutio because it is a ketose (3) \(\alpha-\mathrm{D}(+)\)-glucose and \(\beta-\mathrm{D}(+)\)-glucose are anomers (4) D-glucose and D-mannose are C-2 epimers Which of the following statements are correct? (a) 1,2 and 3 (b) 1,3 and 4 (c) 1,2 and 4 (d) 2,3 and 4

Step-by-Step Solution

Answer

The correct answer is (b) 1, 3, and 4.

Step 1: Analyze Statement 1

The statement "monosaccharides are optically active polyhydroxy carbonyl compounds" is asking if monosaccharides have chiral centers that allow optical activity and if they contain multiple hydroxyl (OH) groups and a carbonyl group (either aldehyde or ketone). Monosaccharides do indeed fit this description, thus statement 1 is correct.

Step 2: Evaluate Statement 2

The statement "fructose does not reduce Fehling's solution because it is a ketose" suggests that fructose's structure as a ketose prevents it from acting as a reducing sugar with Fehling's solution. However, fructose can tautomerize to form an aldose that can reduce Fehling's solution under alkaline conditions, so this statement is incorrect.

Step 3: Consider Statement 3

The statement "\(\alpha-\mathrm{D}(+)\)-glucose and \(\beta-\mathrm{D}(+)\)-glucose are anomers" refers to the different forms of glucose that vary only by the configuration of the hydroxyl group at the anomeric carbon (C-1). These are indeed anomers, making statement 3 correct.

Step 4: Review Statement 4

The statement "D-glucose and D-mannose are C-2 epimers" means that these two sugars differ only at the C-2 position, with all other chiral centers being the same. This is true, so statement 4 is correct.

Step 5: Confirm Correct Answer

Based on the evaluation, statements 1, 3, and 4 are correct. Therefore, the correct choice is (b) 1, 3 and 4.

Key Concepts

Monosaccharides

Monosaccharides are simple sugars that form the building blocks of carbohydrates. They are indeed optically active due to the presence of chiral centers. A chiral center is an atom, particularly a carbon, that has four different groups attached to it, resulting in non-superimposable mirror images, similar to our left and right hands.
Each monosaccharide contains multiple hydroxyl (OH) groups and a carbonyl group, which can either be an aldehyde or a ketone. The presence of these groups explains why they are polyhydroxy carbonyl compounds.
Examples of monosaccharides include:

• Glucose
• Fructose
• Galactose

Their optical activity is a crucial feature, making them essential for various biological processes and industrial applications.

Reducing Sugars

Reducing sugars are carbohydrates that can act as reducing agents. This property comes from the potential of their aldehyde or ketone groups to be oxidized.
Fructose is a ketose but can tautomerize, which is a chemical transformation that allows it to become an aldose form in an alkaline environment. This shift enables fructose to partake in reactions like the reduction of Fehling's solution. Thus, it acts as a reducing sugar under these conditions.
In general, most monosaccharides, including glucose, are reducing sugars. Their ability to reduce oxidizing agents is significant in chemical tests for identifying carbohydrates.

Epimers

Epimers are stereoisomers that differ only in the configuration around a single chiral carbon. D-Glucose and D-Mannose are C-2 epimers, meaning they differ specifically at the second carbon.
Understanding epimers is important for studying the structural variety and function of carbohydrates. Even a small change at one carbon can lead to significant differences in properties and biological activities.
Key examples of epimers include:

• D-Glucose and D-Mannose
• D-Galactose and D-Glucose

These examples highlight the molecular diversity within carbohydrate groups and their significant implications in biological specificity.

Anomers

Anomers are a specific type of epimer, differing in configuration solely at the anomeric carbon, which is the carbon derived from the carbonyl carbon of the open-chain form of the sugar.
In the case of glucose, this involves the C-1 carbon when glucose adopts its cyclic form.

• In alpha-D(+)-glucose, the hydroxyl group at C-1 is on the opposite side of the ring as the CH2OH group.
• In beta-D(+)-glucose, the hydroxyl group at C-1 is on the same side as the CH2OH group.

Anomerization plays a pivotal role in the dynamic behavior of carbohydrates in solutions. This variance greatly affects how carbohydrates interact with other molecules, impacting their biological functions and applications in chemistry.