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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 44b
Chapter 23, Problem 44b

Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using your mechanism for part (a), show why cytidine and adenosine (for example) are not so readily hydrolyzed. Explain why this stability is important for living organisms.

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Identify the structure of ribonucleosides, such as cytidine and adenosine. These molecules consist of a ribose sugar linked to a nitrogenous base (cytosine or adenine) via a β-N-glycosidic bond.
Recall the mechanism for hydrolysis of glycosidic bonds. In acidic conditions, the hydrolysis typically involves protonation of the glycosidic oxygen, followed by cleavage of the bond to release the sugar and the base. This process is facilitated by the stability of the carbocation intermediate or transition state.
Analyze the electronic environment of the β-N-glycosidic bond in ribonucleosides. The nitrogen atom in the glycosidic bond is less electronegative than oxygen, making the bond less polarized and less susceptible to protonation. Additionally, the nitrogenous base does not stabilize a carbocation intermediate as effectively as oxygen in other glycosidic bonds (e.g., in DNA).
Explain why the resistance to hydrolysis is biologically significant. Ribonucleosides are components of RNA, which plays critical roles in protein synthesis and other cellular processes. Their stability ensures that RNA molecules remain intact under physiological conditions, allowing them to perform their functions effectively.
Conclude by emphasizing that the relatively strong acid required for hydrolysis of ribonucleosides reflects their chemical stability, which is essential for maintaining the integrity of genetic information and proper cellular function in living organisms.

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Key Concepts

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

Nucleoside Structure

Nucleosides consist of a nitrogenous base linked to a sugar (ribose in the case of ribonucleosides). The glycosidic bond between the base and the sugar is crucial for the stability of the nucleoside. In cytidine and adenosine, the specific orientation and electronic properties of the substituents contribute to the strength of this bond, making hydrolysis less favorable under mild conditions.
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Naming Nucleosides and Nucleotides Concept 2

Acid-Base Chemistry

The hydrolysis of nucleosides involves breaking the glycosidic bond, which can be facilitated by protonation in acidic conditions. However, ribonucleosides like cytidine and adenosine have structural features that stabilize the bond against protonation and subsequent cleavage. Understanding the role of pH and the stability of intermediates is essential for explaining their resistance to hydrolysis.
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The Lewis definition of acids and bases.

Biological Significance of Stability

The stability of ribonucleosides is vital for cellular functions, as it ensures the integrity of RNA molecules during various biological processes. This stability prevents premature degradation, allowing for accurate protein synthesis and regulation of gene expression. In living organisms, maintaining the balance of nucleoside stability is crucial for proper cellular function and overall homeostasis.
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Related Practice
Textbook Question

Cellulose is converted to cellulose acetate by treatment with acetic anhydride and pyridine. Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into fibers. Show the structure of cellulose acetate.

Textbook Question

Cytosine, uracil, and guanine have tautomeric forms with aromatic hydroxy groups. Draw these tautomeric forms.

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Textbook Question

An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.

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Textbook Question

Glucose is the most abundant monosaccharide. From memory, draw glucose in

(a) the Fischer projection of the open chain.

(b) the most stable chair conformation of the most stable pyranose anomer.

(c) the Haworth projection of the most stable pyranose anomer.

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Textbook Question

Without referring to the chapter, draw the chair conformations of

(a) β-D-mannopyranose (the C2 epimer of glucose).

Textbook Question

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine here. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in Figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and an electron-rich pair of nonbonding electrons.

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