Chapter 14, Problem 14.44AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as $\alpha$ or $\beta$ isomer.

Concept Introduction:

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.

If the $\text{OH}$ group in the new chiral center is drawn up means, then it is known as $\beta$ isomer.  If the $\text{OH}$ group in the new chiral center is drawn down means, then it is known as $\alpha$ isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

b.

Interpretation Introduction

Interpretation:

Other cyclic form of the monosaccharide given has to be drawn.

Concept Introduction:

Refer part “a.”.