Structural formula for cyclic form for ribose, glucose, fructose, and galactose has to be drawn. 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 contains only one sugar unit. Monosaccharides can be drawn in a linear style. Here, the hydroxyl groups are present on the each carbon atom. Open-chain form of monosaccharide is drawn by placing the hydroxyl groups on right or left considering the chiral center. 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.

Question

Want to see the full answer?

Answer 1

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 20, Problem 8RQE

Interpretation Introduction

Interpretation:

Structural formula for cyclic form for ribose, glucose, fructose, and galactose has to be drawn.

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 contains only one sugar unit.

Monosaccharides can be drawn in a linear style. Here, the hydroxyl groups are present on the each carbon atom. Open-chain form of monosaccharide is drawn by placing the hydroxyl groups on right or left considering the chiral center.

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Consider the following SN 2 reaction: مار + Br H₂O acetone + Br OH What effect would each of the following changes have on the rate of this reaction. Select the single best answer for each part. Part 1 of 3 If the substrate was changed to: The rate would Br O increase O decrease O remain unchanged Part 2 of 3 × S If the nucleophile was changed to OH, the rate would: O increase O decrease O remain unchanged Part 3 of 3 If the solvent was changed to ethanol, the rate would: Increase O decrease O remain unchanged 2 ol Ar

Consider the following nucleophilic substitution reaction. The compound listed above the arrow is the solvent for the reaction. If nothing is listed over the arrow, then the nucleophile is also the solvent for the reaction. Part: 0/2 Part 1 of 2 Br acetone + I What is the correct mechanism for the reaction? Select the single best answer. OSN 1 OSN 2 X Part: 1/2 Part 2 of 2 Draw the products for the reaction. Include both the major organic product and the inorganic product. If more than one stereoisomer is possible, draw only one stereoisomer. Include stereochemistry where relevant. Click and drag to start drawing a structure. Х 5 ☐

Triethyloxonium tetrafluoroborate reacts with ethanol (CH3CH2OH) to give diethyl ether (CH3CH2OCH2CH3). BF triethyloxonium tetrafluoroborate Which equation, including the curved arrows, best represents the rate-determining step in the mechanism? Select the single best answer. O OH CH3CH2 OH + H. 0+ CH₂H₂ :0 + 0+ ж + H + :0: 0 C

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 20, Problem 8RQE

Interpretation Introduction

Interpretation:

Structural formula for cyclic form for ribose, glucose, fructose, and galactose has to be drawn.

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 contains only one sugar unit.

Monosaccharides can be drawn in a linear style. Here, the hydroxyl groups are present on the each carbon atom. Open-chain form of monosaccharide is drawn by placing the hydroxyl groups on right or left considering the chiral center.

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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 20, Problem 8RQE

Interpretation Introduction

Interpretation:

Structural formula for cyclic form for ribose, glucose, fructose, and galactose has to be drawn.

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 contains only one sugar unit.

Monosaccharides can be drawn in a linear style. Here, the hydroxyl groups are present on the each carbon atom. Open-chain form of monosaccharide is drawn by placing the hydroxyl groups on right or left considering the chiral center.

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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 20, Problem 8RQE

Interpretation Introduction