Chapter 20, Problem 8RQE

Interpretation Introduction

Interpretation:

Structural formula for cyclic form for ribose, glucose, fructose, and galactose has to be drawn.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides.  They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal.  Monosaccharides contains only one sugar unit.

Monosaccharides can be drawn in a linear style.  Here, the hydroxyl groups are present on the each carbon atom.  Open-chain form of monosaccharide is drawn by placing the hydroxyl groups on right or left considering the chiral center.

Monosaccharide can be expressed in a cyclic form.  In case of an aldohexose, the hydroxyl group present on the $\text{C5}$ carbon atom reacts with the carbonyl group present in $\text{C1}$ carbon atom resulting in formation of a six-membered ring.  Procedure to be followed for obtaining cyclic structure are given as follows.

• Carbon skeleton has to be rotated to $90°$.  While rotating, the groups that are present on the right side ends up below after rotation.
• Chain has to be twisted in order to put the hydroxyl group closer to the carbonyl group of aldehyde.  The ${\text{CH}}_{\text{2}}\text{OH}$ group present on $\text{C5}$ is drawn up.
• The $\text{OH}$ group on the $\text{C5}$ carbon atom reacts with the aldehyde carbonyl resulting in formation of six-membered ring that has a new chiral center.  Considering the orientation of $\text{OH}$ group that is present on the new chiral center that is formed, two isomers are possible.