Chapter 20, Problem 12RQE

Interpretation Introduction

Interpretation:

Structural formula for cyclic form for sucrose and maltose 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.

Disaccharides are drawn by considering the glycosidic linkage that is formed between the two monosaccharide units.