Problem 16 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Which one of the following is not correct for monosaccharides? (a) They are optically active polyhydroxy carbonyl compounds (b) Fructose is ketose sugar and hence it does not give red precipitate with Fehling's solution (c) \(\alpha\) -D (+) glucose and \(\beta-\mathrm{D}\) ( \(+)\) glucose are anomers (d) Glucose and mannose are epimers

Step-by-Step Solution

Answer

Fructose does give a red precipitate with Fehling's solution. Hence, statement (b) is not correct.

Step 1: Understand Monosaccharides

Monosaccharides are the simplest form of carbohydrates. They are polyhydroxy carbonyl compounds and exhibit optical activity due to their chiral centers. It is essential to understand their properties to determine which statement is not correct.

Step 2: Examine Fructose and Fehling's Solution

Fehling’s solution is used to detect reduce sugars. Sugars that can convert into an aldehyde in alkaline solutions can reduce Fehling's solution to give a red precipitate. Although fructose is a ketose, it can tautomerize to an aldose in alkaline conditions and still reduce Fehling’s solution, giving a red precipitate. Therefore, the statement about fructose not giving a red precipitate is incorrect.

Step 3: Compare Anomers

Anomers are types of stereoisomers that differ at the anomeric carbon, and both -\(D\) glucose and \beta\-\(D\) glucose fit this criterion. Thus, \beta\-\(D\) glucose are anomers.

Step 4: Check Epimers

Epimers are stereoisomers that differ in configuration at any single stereocenter except the anomeric carbon, while glucose and mannose differ at the C-2 carbon, making them epimers.

Key Concepts

Anomers

Monosaccharides such as glucose often exist in cyclic forms. In these forms, a new stereocenter, known as the anomeric carbon, is created. This results in the formation of two distinct structures called anomers. Anomers are stereoisomers that vary at the anomeric carbon. For example, - - consider the two forms of glucose: - \( \alpha \)-D-glucose and - \( \beta \)-D-glucose. The difference between these lies in the position of the hydroxyl group attached to the anomeric carbon.If the hydroxyl group is on the opposite side of the ring oxygen, it is the alpha (\( \alpha \) ) form.If it is on the same side as the ring oxygen, it is the beta ( \( \beta \) ) form.Understanding anomers is crucial as they determine the reactivity and interaction of the sugar molecule in biological systems.

Epimers

Epimers are another type of stereoisomers found in monosaccharides. They are carbohydrates that differ only at one stereogenic center, except for the anomeric carbon. A common example is glucose and mannose. - These two sugars differ at the second carbon (C-2) position. - While glucose has the hydroxyl group on the right, mannose has it on the left. This slight difference is enough to give them different properties and biological functions. Epimers are an important concept in carbohydrate chemistry, as they help understand the diversity and specificity of sugar molecules in nature.

Optical Activity

Optical activity is a property seen in molecules that have chiral centers, which are atoms bonded to four different groups. This causes the molecules to exist in two non-superimposable mirror images, known as enantiomers. - Monosaccharides often display optical activity, as they usually contain multiple chiral centers. - When such sugars are dissolved in solution, they rotate plane-polarized light, which is a key property used to learn about their structure and purity. The direction of rotation is designated as - either dextrorotatory (+) or levorotatory (-). Optical activity is an essential concept in the study of organic compounds, helping scientists predict how these molecules will behave in biological systems.

Fehling's Solution

Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrates. Its primary use is to test for the presence of reducing sugars. Reducing sugars are those that can - reduce a mild oxidizing agent - like the blue Fehling's solution. - Upon heating with a reducing sugar, the - solution turns reddish-brown, due to the formation of a copper(I) oxide precipitate. However, not all sugars can reduce Fehling's solution. - While fructose is a ketose sugar - (similar to glucose, which is an aldose), it can isomerize into glucose or mannose in alkaline conditions, which allows it to reduce Fehling's solution. This aspect is vital for determining whether a substance is a reducing sugar, which can be informative in both laboratory and medical settings.