Chapter 24, Problem 28P

Interpretation Introduction

Interpretation:

The $\text{D-aldohexoses}$ that yield the products formed by the given conditions are to be determined.

Concept introduction:

Reduction of monosaccharides with sodium borohydride reduces the aldehyde group to primary alcohol.

Oxidation of monosaccharides with bromine oxidizes the aldehyde group to carboxylic acid.

Oxidation of monosaccharides with nitric acid oxidizes the aldehyde group as well as the terminal ${\text{-CH}}_{\text{2}}\text{OH}$ to carboxylic acid groups.

Carbohydrates undergo enolization in the open-chain form. The aldehyde oxygen is protonated by the hydrogen from C2, converting it to hydroxyl group and forming a double bond between C1 and C2.

Carbohydrates that differ in configuration only at C2 give the same enediol.

A molecule with more than one chirality center can be optically inactive if it has a plane of symmetry of a center of symmetry.