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Ch. 23 - Carbohydrates and Nucleic Acids
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 23 - Carbohydrates and Nucleic AcidsProblem 61a-d
Chapter 23, Problem 61a-d

Some protecting groups can block two OH groups of a carbohydrate at the same time. One such group is shown here, protecting the 4-OH and 6-OH groups of β-d-glucose.

(a) What type of functional group is involved in this blocking group?
(b) What did glucose react with to form this protected compound?
(c) When this blocking group is added to glucose, a new chiral center is formed. Where is it? Draw the stereoisomer that has the other configuration at this chiral center. What is the relationship between these two stereoisomers of the protected compound?
(d) Which of the two stereoisomers in part (c) do you expect to be the major product? Why?

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Step 1: Analyze the structure of the protecting group shown in the image. The protecting group is a benzylidene acetal, which is formed by the reaction of glucose with benzaldehyde. This functional group blocks two hydroxyl groups simultaneously, specifically the 4-OH and 6-OH groups of β-D-glucose.
Step 2: To form this protected compound, glucose reacts with benzaldehyde under acidic conditions. The reaction involves the formation of an acetal, where the aldehyde group of benzaldehyde reacts with the hydroxyl groups of glucose, creating a cyclic structure.
Step 3: When the protecting group is added to glucose, a new chiral center is formed at the carbon atom of the benzylidene group. This occurs because the benzylidene group introduces asymmetry at the carbon atom where it connects to the glucose molecule.
Step 4: To draw the stereoisomer with the other configuration at the new chiral center, invert the stereochemistry of the benzylidene carbon. The relationship between the two stereoisomers is that they are diastereomers, as they differ in the configuration of one chiral center while the rest of the molecule remains the same.
Step 5: The major product is expected to be the stereoisomer where the benzylidene group adopts the configuration that minimizes steric hindrance. This is because steric factors play a significant role in determining the stability of the product, favoring the configuration with less crowding around the chiral center.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Protecting Groups in Organic Chemistry

Protecting groups are temporary modifications used in organic synthesis to prevent certain functional groups from reacting during a chemical reaction. In carbohydrates, protecting groups can shield hydroxyl (OH) groups, allowing selective reactions to occur at other sites. This strategy is crucial for synthesizing complex molecules where specific functional groups need to be preserved while others are modified.
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Chirality and Stereoisomers

Chirality refers to the property of a molecule that makes it non-superimposable on its mirror image, leading to the existence of stereoisomers. A chiral center, typically a carbon atom bonded to four different substituents, creates two stereoisomers known as enantiomers. Understanding chirality is essential for predicting the behavior of molecules in biological systems, as different stereoisomers can have vastly different properties.
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Functional Groups and Their Reactions

Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. In the context of the question, identifying the functional group in the protecting group is crucial for understanding how it interacts with glucose. The reaction between glucose and the protecting group leads to the formation of a new compound, which may involve nucleophilic substitution or other mechanisms depending on the nature of the functional groups involved.
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Related Practice
Textbook Question

Draw the structures of the following nucleotides.

(a) guanosine triphosphate (GTP)

Textbook Question

When the gum of the shrub Sterculia setigera is subjected to acidic hydrolysis, one of the water-soluble components of the hydrolysate is found to be tagatose. The following information is known about tagatose:

(1) Molecular formula C6H12O6

(2) Undergoes mutarotation.

(3) Does not react with bromine water.

(4) Reduces Tollens reagent to give D-galactonic acid and D-talonic acid.

(5) Methylation of tagatose (using excess CH3I and Ag2O) followed by acidic hydrolysis gives 1,3,4,5-tetra-O-methyltagatose.

(a) Draw a Fischer projection structure for the open-chain form of tagatose.

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Textbook Question

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyldisiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.

(a) The TIPDS group is somewhat hindered around the Si atoms by the isopropyl groups. Which OH is more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen.

(b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected.

(c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

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Textbook Question

Draw the structures of the following nucleotides.

(b) deoxycytidine monophosphate (dCMP)

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Textbook Question

Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically active products?

Textbook Question

Show what product results if the aldopentose formed from degradation of X is further degraded to an aldotetrose. Does HNO3 oxidize this aldotetrose to an optically active aldaric acid?

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