FOUND.OF COLLEGE CHEMISTRY
FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
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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 C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.
  • The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

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Chapter 20 Solutions

FOUND.OF COLLEGE CHEMISTRY

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