Ch. 23 - Carbohydrates and Nucleic Acids

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch. 23 - Carbohydrates and Nucleic Acids

Problem 60a

Chapter 23, Problem 60a

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.

Verified step by step guidance

Step 1: Analyze the molecular formula (C6H12O6) and determine that tagatose is a hexose (a six-carbon sugar) with the general formula of a carbohydrate (CnH2nOn). This indicates it is a monosaccharide.

Step 2: Note that tagatose undergoes mutarotation, which implies it is a reducing sugar and exists in equilibrium between its open-chain form and cyclic forms (α and β anomers). This also suggests the presence of a free aldehyde or ketone group in the open-chain form.

Step 3: Observe that tagatose does not react with bromine water. Bromine water selectively oxidizes aldehydes to carboxylic acids, so the lack of reaction indicates that tagatose is a ketose (contains a ketone group rather than an aldehyde group).

Step 4: Consider the reduction of Tollens reagent to form d-galactonic acid and d-talonic acid. This indicates that tagatose is a reducing sugar and can tautomerize to form an aldehyde group in its open-chain form, which reacts with Tollens reagent.

Step 5: Use the information from methylation (1,3,4,5-tetra-O-methyltagatose) to deduce the hydroxyl group positions. Combine this with the stereochemistry of d-galactose and d-tagatose to draw the Fischer projection of the open-chain form of tagatose, ensuring the ketone group is at C2 and the stereochemistry matches the given data.

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Key Concepts

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

Fischer Projection

The Fischer projection is a two-dimensional representation of a molecule that shows the configuration of its stereocenters. In this format, the vertical lines represent bonds that are oriented away from the viewer, while horizontal lines represent bonds that are coming towards the viewer. This method is particularly useful for visualizing carbohydrates, as it allows for easy identification of the orientation of hydroxyl groups and the anomeric carbon in sugars.

Mutarotation

Mutarotation is the change in optical rotation that occurs when an α-anomer and a β-anomer of a sugar interconvert in solution. This process is significant for sugars like tagatose, as it indicates the presence of an anomeric carbon that can exist in two different configurations. The ability to undergo mutarotation is a key characteristic of reducing sugars, which can affect their reactivity and interactions in biochemical processes.

Reducing Sugars

Reducing sugars are carbohydrates that can donate electrons to other molecules, typically due to the presence of a free aldehyde or ketone group. In the case of tagatose, its ability to reduce Tollens reagent demonstrates its reducing properties, which are crucial for various chemical reactions. Understanding the reactivity of reducing sugars is essential for predicting their behavior in organic reactions and their role in metabolic pathways.

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Reducing Sugars

Related Practice

Textbook Question

Sugar X is known to be a D-aldohexose. On oxidation with HNO₃, X gives an optically inactive aldaric acid. When X is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X.

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Textbook Question

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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Textbook Question

Draw the structures of the following nucleotides.

(a) guanosine triphosphate (GTP)

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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

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?

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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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