Chapter 20, Problem 67P

Interpretation Introduction

(a)

Interpretation:

Whether the given monosaccharide is a D or L sugar needs to be determined.

Concept Introduction:

The configuration stereochemistry of the molecule is represented as D and L enantiomers. In the L isomer of a carbohydrate, the $-\text{OH}$ group is attached to the left hand side of the asymmetric carbon away from the carbonyl group and in the D isomer; the $-\text{OH}$ group is on the right hand side.

All the natural sugars are D-isomers.

Answer to Problem 67P

D-isomer

Explanation of Solution

The given monosaccharide is as follows:

From the structure, the hydroxyl group to the carbon atom away from carbonyl group that is C-5 as hydroxyl group on right hand side thus, it is a D-isomer.

Interpretation Introduction

(b)

Interpretation:

The type of carbonyl and number of atoms in the chain of the given monosaccharide needs to be determined.

Concept Introduction:

Monosaccharides are simplest sugar and basic units of carbohydrates. The monosaccharides are further hydrolyzed to form simpler chemical compounds. The general fromula of carbohydrate is ${\text{C}}_n{\text{H}}_{2n}{\text{O}}_n$ . All monosaccharides are water soluble and some of them ae sweet taste. Some of the monoschharides are glucose, fructose, galactose etc.

Aldehyde group with 6 carbon atom: aldohexose.

Answer to Problem 67P

Aldehyde group with 6 carbon atom: aldohexose.

Explanation of Solution

The given monosaccharide is as follows:

There are 6 carbon atoms present in it and there is an aldehyde group thus, it is an aldohexose.

Thus, the type of carbonyl group is aldehyde and number of atoms in chain is 6 carbon atoms.

Interpretation Introduction

(c)

Interpretation:

The enantiomers of the given monosaccharide need to be determined.

Concept Introduction:

The configuration stereochemistry of the molecule is represented as D and L enantiomers. In the L isomer of a carbohydrate, the $-\text{OH}$ group is attached to the left hand side of the asymmetric carbon away from the carbonyl group and in the D isomer; the $-\text{OH}$ group is on the right hand side.

Answer to Problem 67P

Explanation of Solution

The given monosaccharide is as follows:

From the structure, the hydroxyl group to the carbon atom away from carbonyl group that is C-5 as hydroxyl group on right hand side thus, it is a D-isomer.

The enantiomer of the given monosaccharide will be L-isomer. The enantiomer of the D-isomer will be non-superimposable mirror image of it. Thus, the structure of L-isomer will be as follows:

Interpretation Introduction

(d)

Interpretation:

The chirality centers needs to be labelled.

Concept Introduction:

The molecules in which there is one or more chiral centers are known as chiral molecules. A carbon attached to four different groups or atom is chiral in nature. The chiral molecules which are mirror images of each other are known as enantiomers. The chiral molecules are non-superimposable to each other. Non-superimposible means one molecule cannot be placed on the other molecule.

Answer to Problem 67P

The 4 chiral centers are represented as follows:

Explanation of Solution

The given monosaccharide is as follows:

The chiral carbon atom has 4 different groups attached to it. Thus, the labelled carbon atoms are chiral in nature.

Thus, there are 4 chiral centers in the molecules.

Interpretation Introduction

(e)

Interpretation:

The $\alpha$ anomer of the cyclic form of the given monosaccharide needs to be determined.

Concept Introduction:

An epimer is a stereoisomer having difference in configuration at any one chiral center. In the case of anomer, the difference in configuration takes place at the hemiacetal carbon in the cyclic form. The carbon atom is known as anomeric carbon.

Both the $\alpha$ and $\beta$ anomer structures are identical, the only difference is that in $\alpha$ form, the −OH group at C-1 position is down and in $\beta$ form, it is up. They are anomers to each other.

Answer to Problem 67P

Explanation of Solution

The given monosacchride is as follows:

The $\alpha$ isomer will have OH group at C-1 carbon on the right hand side. Thus, the $\alpha$ isomer is represented as follows:

Interpretation Introduction

(f)

Interpretation:

The product from the reaction of monosaccharide with Benedict's reagent needs to be determined.

Concept introduction:

Benedict's reagent is used as a mild oxidizing agent and can oxidize aldehyde into the corresponding acid.

Answer to Problem 67P

Explanation of Solution

The given monosacchride is as follows:

The Benedict's reagent is used as a mild oxidizing agent. Here, Cu²+ complexed with citrate and OH- are present in the Benedict's reagent which is used to identify reducing sugars. It is used as a mild oxidizing agent and can oxidize aldehyde into the corresponding acid. A hydroxyl group is added to the carbon with the carbonyl group when this reagent is added. Cu²+ is converted to Cu+ which then precipitates as Cu₂O which is a brick red precipitate.

The reaction is represented as follows:

Interpretation Introduction

(g)

Interpretation:

The product formed on reaction of monosaccharide with ${\text{H}}_{\text{2}}$ and Pd needs to be determined.

Concept introduction:

H₂, Pd is used as a reducing agent to reduce compounds like alkenes, alkynes, aldehydes, ketones etc. The reduction of aldehyde results in the formation of alcohol.

Answer to Problem 67P

Explanation of Solution

When H₂, Pd is added to aldehyde (reducing sugars), the corresponding alcohol is formed. The double bonded oxygen of the aldehyde turns to a hydroxyl group and one hydrogen atom is added to the carbon containing the respective oxygen atom. Pd is the catalyst of this reaction. H₂ breaks and gets added as H atoms to oxygen atom and the carbon atom.

The given monosacchride is as follows:

Thus, the reaction with H₂, Pd is represented as follows:

Interpretation Introduction

(h)

Interpretation:

The monosaccharide needs to be labelled as a reducing or non-reducing sugar.

Concept Introduction:

A reducing sugar is defined as any sugar capable of behaving as a reducing agent having a free aldehyde group or a free ketone group. All monosaccharide are reducing sugars. There are some disaccharides, oligosaccharides and polysaccharides which are also reducing sugars.

Answer to Problem 67P

The given monosaccharide is a reducing sugar.

Explanation of Solution

The given monosaccharide is represented as follows:

Since, all the monosaccharides are reducing sugar thus, it is also reducing sugar. A reducing sugar acts as a reducing agent. Thus, on reaction with ${\text{Cu}}^{2+},{\text{OH}}^{-}$ the monosaccharide gets oxidized as follows: