Chapter 14, Problem 14.21UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharides has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

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 that have the carbonyl group at the terminal carbon atom $\text{C1}$ are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom $\text{C2}$ are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide.  They are given below.

• Carbon chain with three carbon atoms is triose.
• Carbon chain with four carbon atoms is tetrose.
• Carbon chain with five carbon atoms is as pentose.
• Carbon chain with six carbon atoms is as hexose.

Explanation of Solution

Structure of monosaccharide A is drawn as shown below.

The carbonyl group is present on the terminal carbon atom $\text{C1}$.  Thus, this is an aldose.  The total number of carbon atoms present in the chain is four.  Thus, this monosaccharide is a tetrose.  Considering all these, the monosaccharide A is classified as an aldotetrose.

Structure of monosaccharide B is drawn as shown below.

The carbonyl group is present on the terminal carbon atom $\text{C1}$.  Thus, this is an aldose.  The total number of carbon atoms present in the chain is six.  Thus, this monosaccharide is a hexose.  Considering all these, the monosaccharide B is classified as an aldohexose.

b.

Interpretation Introduction

Interpretation:

Chiral centers has to be located in the given compounds.

Explanation of Solution

Structure of monosaccharide A is drawn as shown below.

The two carbon atoms that are present in the middle is found to be bonded with four different groups.  Thus, there are two chiral centers present in monosaccharide A.  This is depicted as follows.

Structure of monosaccharide B is drawn as shown below.

The four carbon atoms that are present in the middle is found to be bonded with four different groups.  Thus, there are four chiral centers present in monosaccharide B.  This is depicted as follows.