1D & 5D - Carbohydrates (BC, OC)

Question 1

A 6 carbon carbohydrate with a functional group aldehyde is named …

Answer 1

Aldohexoses

Question 2

A 5 carbon sugar with a ketone group is called?

Answer 2

Ketopentose

Question 3

What is the most simple aldose? Describe its structure

Answer 3

This is the aldotriose, common name = glyceraldehyde
Note that the each C of a carbohydrate has an attachment to OH and the first carbon is a carboxyl carbon with a functional group aldehyde.

Question 4

Describe the most simple ketose.

Answer 4

The most simple ketose is the dihydroxyacetone

Structure: 3 C, each C (C1 and C3) are bound to an OH while the C2 is a carbonyl double bonded to O (ketone)

Question 5

The nomenclature of carbohydrates are dependent on what aspects?

Answer 5

The name of a carbohydrate is based on its content!

1. # of carbs
2. # of sugar moieties
3. # of of functional groups
4. Stereochemistry of sugar

Question 6

Determine the relationship between D-erythrose and L-erythrose in terms of stereochemistry.

Answer 6

Both are mirror images of one another and therefore non superimposable to one another. As a result, these 2 molecules are enantiomers to one another

Question 7

What is the stereochemical relationship between D-erythrose and D-threose?

Answer 7

Both are not mirror images because they differ at one attachment. This means they are nonsuperimposable. As a result, they are diastereomers, more specifically epimers because they differ in attachment at one carbon

Question 8

Explain the stereochemistry seen in carbohydrates

Answer 8

Carbohydrates are optical compounds and therefore they have different stereocenters seen within their C centers. Almost all of the C centers in a sugar compound have different attachments and have an attachment to an OH, H and 2 different C
In biochemistry, optical activity is labeled as either D and L (in O chem, optical activity is termed R and S)
D configurations have a positive rotation [S enantiomers are Dextrorotatory therefore positive and clockwise]
L configurations have a negative rotations [R enantiomers are Levorotatory therefore negative and counter-clockwise]

Question 9

Q - Where on the fischer projection of S-glyceraldehyde do the OH point to?

Answer 9

Remember that S configurations are D configurations! Positive configurations have their priority groups of stereocenters point to the right!
Therefore OH of S-glyceraldehyde point to the right on a fischer projection

Question 10

You’re drawing a fischer projection of a monosaccharide and you note that your compound is a L-monosaccharide. How should you draw the OH on the stereocenters

Answer 10

L sugars have OH to the left of the fischer. You should only represent stereocenters in the intersections of the fischer projection
Remember that L configurations are R configurations as well!

Question 11

In representing a molecule on a fischer projection, where does the C1 begin? Use glucose as an example

Answer 11

C1 starts on the top of page and because it does not have a stereocenter, this means it does not have a intersection. With glucose, its aldehyde group will be here
Intersections represents the stereo carbon with different attachments
Last carbon of the chain is at the bottom of the page

Question 12

What is the stereocenter configuration of the carbon center when you place the highest priority molecule to the right on a fischer projection. What about on the left?

Answer 12

This is in reference to the local configuration of each center.
When placed on the right, the configuration is R
When placed on the left, the configuration is S

Question 13

Draw the less stable anomer of D - Glucose in Hawthorne projection form

Answer 13

Less stable anomers should have their OH equatorial to the CH2OH AKA cis. This molecule is the ß - OH

Question 14

Convert a a D glucose from fischer to Haworth.

Answer 14

1. Count the number of carbons in the straight chain from carbonyl carbon to nucleophilic O - this is the number of carbons that will be the base of your cyclic structure (which is the basis of haworth - representation of 3D SUGAR cyclic structure)
2. Draw Haworth base. O is part of the cyclic ring. D sugars - C6 is going to point UP and L sugars - C6 are going to look down
3. Right groups on fischer projection project bottom in haworth
4. Left groups on fischer projection project up in haworth

Question 15

Haworth projection is a model used to depict the 3D structure of cyclic sugars. Describe how 5C rings are different.

Answer 15

In reality 5C rings are planar compared to 6C rings tend to exist in chair conformation

Question 16

As an aldose converts from a straight chain into a cyclic form its name changes. What is this name change? How do ketones name also change?

Answer 16

Aldoses => hemiacetals

Ketoses => hemiketals

Question 17

All of the following are able to contribute to causes your hemiacetal to mutarotate except: 
A. Water 
B. Glycosylase
C. Acid
D. Base

Answer 17

D. Glycosylase. This is an enzyme that repairs DNA
Mutarotation is the process in which sugar molecules are able to rotate from alpha to beta structures. This process is initiated by solutions only of water, acid and base. This process occurs, by temporarily opening the ring, and having the same O attack the carbonyl carbon. In this reattack to close the cyclic ring, the structure has one of 2 choices to create an alpha or beta ring

Question 18

Which structure is the most stable structure of cyclic glucose?

Answer 18

α-glucose is more stable than ß-glucose. This is because α-OH is in the axial form and this creates steric strain. Because of this, ⅓% of sugars are only found in this configuration while the other ⅔% of sugars are found to be in ß-glucose

Question 19

Which of the diastereomers of glucose from the previous question are considered to be epimers of glucose? Enantiomers?

Answer 19

Epimers of D-glucose - D-Allose, D-allose & D-galactose
Enantiomers - none of D stereoisomers are enantiomers of glucose, but L-glucose is an enantiomer of D-Glucose (beyond the question)