Organic Chemistry : Help with Organic Carbohydrates

Study concepts, example questions & explanations for Organic Chemistry

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Example Questions

Example Question #1 : Help With Organic Carbohydrates

Chemists and biochemists have many ways of representing sugars. Glucose, the most common hexose, is shown below in various linear and cyclic projections. Using the linear and cyclic projection of your choice, can you indicate which colored oxygen in the linear form corresponds to the circled hemiacetal oxygen once the cyclization reaction is complete?

Q3

Possible Answers:

Blue

Purple

Green

Yellow

Red

Correct answer:

Purple

Explanation:

This answer, regardless of your preference of projection type, is easiest to obtain using arrow pushing for the cyclization reaction to keep track of each carbon and oxygen:

A3

The purple carbon in the linear projection ends in the circled hemiacetal position.

Example Question #2 : Identifying Monosaccharides

Which of the following structures represents the anomeric alpha ring structure of D-glucose? 
Linear glucose

Possible Answers:

Alpha ribose

Alpha galactose

Alpha glucose

Alpha mannose

Alpha fructose

Correct answer:

Alpha glucose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-glucose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be trans, cis, and trans with respect to the .

Example Question #3 : Identifying Monosaccharides

Which of the following structures represents the anomeric alpha ring structure of D-galactose? 

Linear galactose

Possible Answers:

Alpha fructose

Alpha galactose

Alpha glucose

Alpha mannose

Alpha ribose

Correct answer:

Alpha galactose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-galactose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be trans, cis, and cis with respect to the .

Example Question #4 : Identifying Monosaccharides

Which of the following ring structures represents the anomeric alpha ring structure of D-mannose? 

Linear mannose

Possible Answers:

Alpha ribose

Alpha fructose

Alpha mannose

Alpha glucose

Alpha galactose

Correct answer:

Alpha mannose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-mannose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be cis, cis, and trans with respect to the .

Example Question #2 : Help With Organic Carbohydrates

Identify the aldose pictured, including its alpha or beta designation. 

A d mannose

Possible Answers:

-D-glucose

-D-mannose

-D-ribose

-D-mannose

-D-glucose

Correct answer:

-D-mannose

Explanation:

The structure pictured is mannose because the hydroxyl groups at carbons 2, 3, and 4 are situated cis, cis, and trans (respectively) to the  attached to carbon 5.

The mannose pictured is in alpha form because the hydroxyl group at carbon 1 is trans to the  attached to carbon 5.

Example Question #3 : Help With Organic Carbohydrates

The Fischer projection pictured is a form of glucose. The carbon labeled "x" is the chiral carbon farthest away from carbon 1 and the hydroxyl group connected to carbon "x" is on the right. This fact designates that the glucose as what configuration? 

D configuration

Possible Answers:

Alpha

L

D

Beta

Pyranose

Correct answer:

D

Explanation:

The chiral carbon farthest away from carbon 1 is designated as "D" if its hydroxyl group is on the right side in the Fischer projection. In other words, this is D-glucose because the hyroxyl group on carbon "x" is oriented to the right.

Example Question #4 : Help With Organic Carbohydrates

What is the name of the aldose pictured in this Fischer projection?

D ribose

Possible Answers:

L-xylose

L-lyxose

L-fructose

D-ribose

D-arabinose

Correct answer:

D-ribose

Explanation:

The structure is D-ribose because it is a five-carbon aldose with the hydroxyl groups on carbons 2, 3, and 4 all on the right in the Fischer projection.

Example Question #3 : Help With Organic Carbohydrates

Which of the following statements is true regarding carbohydrates?

Possible Answers:

The anomeric carbon is the site of attachment from one monosaccharide to another

Aldoses are more common in nature than ketoses

In nature, carbohydrates are usually found with a "D" conformation

All of these

Correct answer:

All of these

Explanation:

All of these statements are true. A carbohydrate is said to have a "D" conformation in its acyclic form when the alcohol group on the carbohydrate's top stereocenter is on the right side in a Fischer projection. Most carbohydrates that we deal with in organic chemistry are aldoses, which means that they contain an aldehyde. The anomeric carbon is the site of attachment from one monosaccharide to another, and can be used to create polysaccharides.

Example Question #5 : Help With Organic Carbohydrates

What is the name of the following carbohydrate?

Screen shot 2015 11 16 at 12.44.13 pm

Possible Answers:

Alpha-L-galactopyranose

Alpha-L-glucopyranose

Alpha-D-galactopyranose

Beta-D-galactofuranose

Beta-D-galactofuranose

Correct answer:

Alpha-D-galactopyranose

Explanation:

Stereochemistry from second to fifth carbon is R, S, S, R, which indicates D-galactose. The Haworth structure is a six-membered ring, so the molecule is in its pyranose form. The molecule has its anomeric hydroxyl group pointing down, so it's the alpha anomer.

Example Question #6 : Help With Organic Carbohydrates

What is the name of the following molecule?

Screen shot 2015 11 16 at 12.55.54 pm

Possible Answers:

Alpha-D-glucopyranose

Alpha-D-glucofuranose

Beta-D-galactofuranose

Alpha-L-galactopyranose

Beta-L-glucofuranose

Correct answer:

Alpha-D-glucofuranose

Explanation:

Stereochemistry from second to fifth carbon is R, S, R, R, which indicates D-glucose. The Haworth structure is a five-membered ring, so the molecule is in its furanose form. The molecule has its anomeric hydroxyl group pointing down, so it's the alpha anomer.

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