Question

The structural formula for the linear form of D-mannose is (a) How many chiral carbons are present in the molecule? (b) Draw the structure of the six-member- ring form of this sugar.

Short answer

(a) D-Mannose has $5$ chiral carbons in its linear form. (b) The six-membered ring form of D-Mannose is formed by an internal redox reaction where the oxygen on the C5 hydroxyl group attacks the C1 carbonyl carbon, creating a hemiacetal intermediate. This leads to a six-membered ring structure, where the -OH groups on the left are drawn down and the -OH groups on the right are drawn up, except for the -OH on the new chiral carbon C1 (alpha or beta configuration). The -CH2OH group attached to C5 is drawn up.

Step-by-step solution

Identify chiral carbons

Chiral carbons, also known as asymmetrical carbons, are carbons attached to four different types of groups. In the case of D-Mannose, we are looking at a structure with six carbons. Starting from the bottom of the molecule (the aldehyde group), we ignore the bottom Carbon (C1), as it's not a chiral carbon; it's attached to two hydrogen atoms (one directly and another through the aldehyde group). We then look at the other carbons (from C2 to C6), each attached to -H, -OH, -CH2OH and another carbon atom with different groups. These are all chiral carbons.

Count chiral carbons

Now that we know what chiral carbons are, we can count them in the molecule. We already established that the bottom carbon is not chiral, but the remaining five are, making for a total of five chiral carbons in the D-Mannose molecule.

Understand the cyclization process

In order to draw the six-membered ring form, we have to understand how the cyclization process works. In the case of D-Mannose, the molecule undergoes an internal redox reaction where the oxygen on the C5 hydroxyl group (-OH) attacks the C1 carbonyl carbon, creating a hemiacetal intermediate. This process leads to the formation of a six-membered ring due to the five carbon atoms and one oxygen atom.

Draw the six-membered ring form

Start by placing the oxygen atom at the top of the ring, then add the five carbon atoms around the ring. Remember that this is a Haworth projection, so the atoms and bonds are not drawn to scale. Next, add the side groups (-H and -OH) to the carbons. The -OH groups on the left in the linear form are drawn below the plane of the ring (down) and those on the right are drawn above the ring (up) except for the -OH on the newly formed chiral carbon C1 which is determined depending on whether it’s in alpha or beta configuration. As it's a D-sugar, the -CH2OH group attached to C5 is drawn up. Please follow standard chemistry books or online resources to draw the correct structural formula.

Key concepts

D-Mannose Structure

Understanding the structure of D-Mannose is essential for comprehending its behavior in biological systems. D-Mannose is a simple sugar, or monosaccharide, with the molecular formula C₆H₁₂O₆. Its structure consists of a chain of six carbon atoms where each carbon atom (except for the first and the last) is attached to one hydrogen atom and one hydroxyl group (-OH), thus making them chiral.

Chiral carbons are those that have four different groups attached to them, making it possible for isomers—compounds with the same molecular formula but different arrangements—to exist. In D-Mannose, there are five chiral carbons, indicated from C2 to C6. Unique to its structure is the aldehyde group at the first carbon (C1), which plays a critical role in the sugar's reactivity and cyclization—a process where the sugar transforms from a linear to a cyclic form.

Cyclization of Sugars

Cyclization is a chemical reaction during which a linear sugar molecule, like that of D-Mannose, reacts to form a ring structure. This transformation is a result of the reaction between functional groups within the same molecule.

In D-Mannose, cyclization occurs when the oxygen on the hydroxyl group of the fifth carbon (C5) attacks the carbonyl carbon (C1), forming a new chiral center at C1 and resulting in a cyclic hemiacetal. This process yields a six-membered ring known as a pyranose, resembling the structure found in many naturally occurring sugars. The orientation of the hydroxyl groups around the ring is crucial for the sugar's properties and will influence such factors as its taste, reactivity, and recognition by enzymes.

Haworth Projection

After sugars cyclize, they are often depicted using Haworth projections. A Haworth projection is a common way to represent cyclic sugar molecules in two dimensions while still giving an impression of their three-dimensional form.

The projection shows a simple hexagon for a six-membered sugar ring such as D-Mannose. Atoms that are on the right in the linear form of the sugar point up in the ring form, and those on the left point down. However, for the hydroxyl group on the newly formed chiral center at C1, the direction can vary, indicating whether the sugar has an alpha or beta configuration. The positioning of these hydroxyl groups affects how the sugar interacts with other molecules, making the Haworth projection a valuable tool for visualizing and understanding sugar chemistry.

Hemiacetal Formation

Hemiacetal formation is a key step in the cyclization of sugars such as D-Mannose. A hemiacetal is formed when an aldehyde reacts with an alcohol. For sugars, this happens intramolecularly when an aldehyde at C1 reacts with a hydroxyl group (-OH), resulting in a carbon atom that's simultaneously connected to an -OH group and an ether (R-O-R') linkage.

In the specific case of D-Mannose, the hemiacetal forms when the hydroxyl group on C5 attacks the aldehyde at C1, creating a new chiral center. The formation of a hemiacetal is reversible, leading to an equilibrium between the open-chain and the cyclic form of the sugar in solution. This reversible process is critical for many biological reactions involving carbohydrates, especially those related to energy metabolism and structural components in cells.