Chapter 14, Problem 14.25UKC

a.

Interpretation Introduction

Interpretation:

Given monosaccharide is a D or L sugar has to be shown.

Concept Introduction:

A carbon atom that is bonded to four different groups is known as a chiral carbon atom.  This can rotate the plane polarized light.  D- and L- isomers of monosaccharide can be identified by looking into the chiral center that is farther from the carbonyl group.  In a Fischer projection, if the $\text{OH}$ group is present on the right side in the chiral center that is farthest from the carbonyl group means it is a D monosaccharide and if the $\text{OH}$ group is present on the left side means it is a L monosaccharide.

Explanation of Solution

Given monosaccharide structure is:

This contains a carbonyl group in the terminal carbon atom $\text{C1}$.  The hydroxyl group present in the chiral center that is farthest from the carbonyl group is on the right side.  Hence, this monosaccharide is classified as D-sugar.

b.

Interpretation Introduction

Interpretation:

The given monosaccharide has to be classified based on the number of atoms in chain and 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

Given monosaccharide is:

This contains a carbonyl group in the terminal carbon atom.  The total number of carbon atoms in the carbon chain is found to be six.  Therefore, the given monosaccharide is classified as an aldohexose.

c.

Interpretation Introduction

Interpretation:

Chiral centers has to be labelled in the given monosaccharide.

Explanation of Solution

Given monosaccharide is:

Chiral carbon is the one that contains four different groups bonded to the same carbon atom.  There are four chiral centers in the given monosaccharide.  The chiral centers are labelled as follows.

d.

Interpretation Introduction

Interpretation:

Enantiomer has to be drawn for the given monosaccharide.

Concept Introduction:

Enantiomers are two stereoisomers of a compound that rotate the plane polarized light exactly opposite.  The configuration present in the enantiomers will be exactly opposite to each other.

Explanation of Solution

Given monosaccharide is:

The given monosaccharide is found to be a D-monosaccharide.  Therefore, the enantiomer of this has to be L-monosaccharide.  The structure of L-monosaccharide can be obtained by drawing the mirror image of the D-monosaccharide.

e.

Interpretation Introduction

Interpretation:

The $\alpha$ isomers has to be drawn for given monosaccharide.

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.

Explanation of Solution

Given monosaccharide structure is.

This is D monosaccharide.  First step is to rotate the monosaccharide to $90°$.  This is done as shown below.

Second step is to twist the chain in order to put the $\text{OH}$ group on the $\text{C5}$ carbon atom to be near the carbonyl group.

Reaction between the $\text{OH}$ group and carbonyl group results in formation of two isomers.  They are $\alpha$ isomer and $\beta$ isomer.  $\alpha$-isomer is the one that has the hydroxyl group on the carbon atom $\text{C1}$ drawn down.

f.

Interpretation Introduction

Interpretation:

Product that will be formed when the given monosaccharide is treated with Benedict’s reagent has to be drawn.

Concept Introduction:

Monosaccharides that are aldoses contains an aldehyde group.  The carbonyl of the aldehyde undergoes oxidation very easily resulting in the formation of carboxyl group.  The product formed is known as aldonic acid.  Benedict’s reagent is ${\text{Cu}}^{\text{2+}}$.  The monosaccharides that contains aldehyde group are known as reducing sugars.  In the course of the reaction, Benedict’s reagent is reduced to ${\text{Cu}}^{\text{+}}$.  General reaction can be represented as follows.

    ${\text{Aldose + 2Cu}}^{\text{2+}}\to \text{Aldonic}{\text{acid + Cu}}_{\text{2}}\text{O}$

Explanation of Solution

Given monosaccharide is:

This contains an aldehyde carbonyl group.  Thus, this is an aldose.  When this is treated with Benedict’s reagent, the aldehyde group is converted into carboxyl group by oxidation.  The complete reaction can be represented as follows.

g.

Interpretation Introduction

Interpretation:

Product that is formed when given monosaccharide is treated with hydrogen in presence of palladium has to be given.

Concept Introduction:

Carbohydrates undergo reduction similar to an alkene.  When an aldose is treated with hydrogen in presence of palladium as catalyst the carbonyl group present in the aldose is reduced resulting in formation of alditol.  This is also known as “sugar alcohol”.

Explanation of Solution

Structure of given monosaccharide is.

When this is treated with hydrogen in presence of palladium, the carbonyl group is changed into alcohol.  The product and the complete reaction is depicted as shown below.