The monosaccharide given has to be labelled as α or β isomer. Concept Introduction: Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH 2 OH group present on C5 is drawn up. The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible. If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer. This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections . The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

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Answer 1

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Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

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Answer 2

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

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Answer 3

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

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Answer 4

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

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Answer 5

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

Answer 6

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

Answer 7

Chapter 14, Problem 14.45AP

a.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Monosaccharide can be expressed in a cyclic form. In case of an aldohexose, the hydroxyl group present on the C5 carbon atom reacts with the carbonyl group present in 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 CH2OH group present on C5 is drawn up.

The OH group on the 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 OH group that is present on the new chiral center that is formed, two isomers are possible.

If the OH group in the new chiral center is drawn up means, then it is known as β isomer. If the OH group in the new chiral center is drawn down means, then it is known as α isomer.

This cyclic form of representation of the monosacccharides as flat, six-membered rings is known as Haworth projections.

The same rule applies for five membered ring also if it is formed from aldopentose and ketohexose.

Answer 8

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

Answer 9

b.

Interpretation Introduction

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 14.1 - Draw a Lewis structure for glucose that clearly...Ch. 14.2 - Prob. 14.2P Ch. 14.2 - Prob. 14.3P Ch. 14.2 - Prob. 14.4P Ch. 14.2 - Rank the following compounds in order of...Ch. 14.2 - Prob. 14.6P Ch. 14.2 - Prob. 14.7P Ch. 14.2 - Prob. 14.8P Ch. 14.2 - Prob. 14.9P Ch. 14.3 - Prob. 14.10P

Solution Summary: The author explains that monosaccharide can be expressed in a cyclic form.

Concept explainers

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.

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Chapter 14 Solutions

Solution Summary: The author explains that monosaccharide can be expressed in a cyclic form.

Concept explainers

Interpretation: The monosaccharide given has to be labelled as α or β isomer. Concept Introduction: Refer part “a.”.

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Chapter 14 Solutions

Interpretation:

The monosaccharide given has to be labelled as α or β isomer.

Concept Introduction:

Refer part “a.”.