Question

The structural formula for the linear form of galactose is O=CC(O)C(O)C(O)C(O)CO (a) Is this molecule a sugar? (b) How many chiral carbons are present in the molecule? (c) Draw the structure of the sixmember-ring form of this molecule.

Short answer

(a) Yes, this molecule is a sugar. (b) There are 3 chiral carbons in the molecule. (c) The six-member-ring form of the molecule is: ``` O | HO-3-C-6-CH2OH | 3-C | HO-4-C-H | O ```

Step-by-step solution

Identify if the given molecule is a sugar

The given molecular formula for the linear form of galactose is: \begin{equation} O=CC(O)C(O)C(O)C(O)CO \end{equation} To check if this molecule is a sugar, we need to see if it's in the empirical formula of \(C_n(H_2O)_n\). The molecule contains six carbons, six hydroxyl groups (OH) bonded to carbons, and an aldehyde group (H-C=O). Therefore, it is in the empirical form \(C_6(H_2O)_6\), which confirms it's a sugar. (a) Yes, this molecule is a sugar.

Count the chiral carbons in the molecule

To count the chiral carbons in the molecule, we need to find the carbon atoms bonded to four different groups. In the given molecular structure, the carbons bonded to hydroxyl groups are chiral carbons. Excluding the first carbon (since it's part of the aldehyde group) and the last carbon (since it's bonded to only 3 different groups): - Carbon 2 is bonded to OH, H, and 2 other carbons: chiral - Carbon 3 is bonded to OH, H, and 2 other carbons: chiral - Carbon 4 is bonded to OH, H, and 2 other carbons: chiral - Carbon 5 is bonded to OH, H, and 1 other carbon (and a hydrogen): not chiral (b) There are 3 chiral carbons in the molecule.

Draw the six-member-ring form of the molecule

To draw the six-member-ring form, we form a ring by bonding carbon 1 and carbon 5 via an oxygen atom. Also, ensure that the orientation of the OH groups remains the same. The hydroxyl groups on carbons 2, 3, and 4 fall below the plane of the ring, while the hydroxyl groups on carbon 1 and carbon 5 remain above the plane. The resulting structure in Haworth projection is: ``` O | HO-3-C-6-CH2OH | 3-C | HO-4-C-H | O ``` (c) The six-member-ring form of the molecule is drawn above.

Key concepts

Sugar Chemistry

Sugar chemistry is a fascinating and vital area of biochemistry, focusing on the study of various types of sugars and their chemical properties. Sugars are organic molecules known as carbohydrates, essential for life as they provide energy to living organisms, form structural components in cells, and are involved in numerous biological processes.

At the most basic level, sugars are classified as monosaccharides, disaccharides, and polysaccharides. Monosaccharides like glucose and galactose are simple sugars with a general empirical formula of \(C_n(H_2O)_n\), where \(n\) is usually a number between three and seven. These simple sugars are the building blocks for more complex carbohydrates.

Galactose, the sugar under examination in the exercise, is a monosaccharide comprising six carbon atoms, twelve hydrogen atoms, and six oxygen atoms, following the general formula for carbohydrates ( \(C_6H_{12}O_6\) ). As the first step of the solution confirms, the structure provided does indeed correspond to a sugar molecule following this generic formula.

Chiral Carbons

Chiral carbons are a central concept in organic chemistry. A carbon atom is considered chiral if it is attached to four different chemical groups, making it asymmetrical. This lack of symmetry allows for the molecule to exist in two different forms, known as enantiomers, which are non-superimposable mirror images of each other.

Chirality is crucial in biochemistry because it determines how biological molecules interact with one another. Often, only one enantiomer is biologically active and can have drastically different effects compared to its mirror image. In sugars, the configuration of chiral carbons defines the sugar's specific structure and properties.

In the exercise, students learn to identify chiral carbons within the linear structure of galactose, where only the carbon atoms bonded to four different groups are considered chiral. According to the solution, galactose contains three such chiral centers, excluding the terminal carbons which do not meet the criteria for chirality.

Haworth Projection

The Haworth projection is a simplified way to depict the cyclic forms of sugars, which are more prevalent in nature than their linear forms. This representation was named after the British chemist Sir Walter Norman Haworth.

A Haworth projection is a diagrammatic depiction that shows the cyclic sugar structure as if it were flat, with the ring typically drawn in a hexagonal or pentagonal shape for six-membered or five-membered rings, respectively. The orientation of groups attached to the ring carbon atoms indicates whether they point above or below the ring plane.

In the context of galactose, which is represented in the exercise solution, the Haworth projection provides a more realistic view of sugar's structure in its cyclic form. The solution's step-wise approach assists students in visualizing the transformation from the linear structure to the six-membered ring, showing the position of the hydroxyl (OH) groups in relation to the plane of the ring, which helps in understanding the molecule's three-dimensional conformation.