Chapter 23, Problem 62A

Interpretation Introduction

Interpretation:

The structure of the open chain from of fructose should be drawn. All chiral carbons should be circled and number of stereoisomers should be calculated for fructose.

Concept introduction:

The compounds having similar chemical formula but different structures are known as isomers.

The compounds having similar chemical or molecular formula but different connectivity is known as constitutional isomers.

Number of stereoisomers is calculated by below expression:

${\text{2}}^n$

Where, n = number of chiral carbons

Answer to Problem 62A

The open chain structure of fructose with highlighted chiral carbons is:

Number of isomers = 8

Explanation of Solution

Fructose consists of single unit of sugar which can’t be hydrolyzed into simpler molecule.

When a carbon atom is linked with different groups is known as chiral carbon and center is known as chiral center. Chiral center is represented by a symbol *.

The structure is drawn as with symbol *:

Thus, three chiral centers are present in the above structure as first three carbon atoms are linked with four different groups.

Number of isomers is calculated as:

Here, n = 3 (3 chiral carbon atoms are present)

Number of isomers = ${\text{2}}^3$

= 8

Therefore, eight isomers are present for the structure of fructose.