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

(b) 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.

Short Answer

Expert verified

a)

b)

Nucleosides require stronger acid, or higher temperatures to be hydrolyzed. This is important in living systems as it would cause genetic damage to an organism if its DNA or RNA were too easily decomposed.

Step by step solution

01

Step-1. Mechanism of acid-catalyzed hydrolysis of an aliphatic riboside:

An aliphatic riboside on acid-catalyzed hydrolysis produces hemiacetal form of ribose. Tertiary aliphatic amine which is present in the aliphatic riboside acts as a strong base and abstracts proton from the medium which leads to formation of positive charge on nitrogen atom, which is not stable, thus, the tertiary amine group leaves as a leaving group and carbocation is formed. This carbocation is stable due to mesomeric effect of nearby oxygen and then water acts as a nucleophile and attacks at carbocation. Then, after stabilisation of positive charge on oxygen, we get our required product.

Mechanism of acid-catalyzed hydrolysis of an aliphatic riboside

02

 Step-2. Importance of stability of nucleosides in living systems:

Nucleosides are less rapidly hydrolyzed in aqueous acid because the site of protonation that is, nitrogen in adenosine and oxygen atom with negative charge in cytidine is much less basic than aliphatic amine in aminoglycoside. Nucleosides require longer time, or stronger acid, higher temperature to be hydrolyzed. This stability is important in living systems as it would cause genetic damage or even death of the organism if decomposition of DNA or RNA takes place easily. Considerable energy is expended to maintain the structural integrity of DNA.

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Most popular questions from this chapter

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

(a) The TIPDS group is somewhat hindered around the Siatoms by the isopropyl groups. Which OHis 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.

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 CH3 I 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.

(b) Draw the most stable conformation of the most stable cyclic hemiacetal form of tagatose.

Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of α-D-galactopyranoseis +150.70 , and that of the β anomer is +52.80 . When either of the pure anomers dissolves in water, the specific rotation gradually changes to +80.20. Determine the percentages of the two anomers present at equilibrium.

The relative configurations of the stereoisomers of tartaric acid were established by the following synthesis:

(1) D-(+)-glyceraldehydediastereomers A and B (separated)

(2) Hydrolysis of A and B using aqueous Ba(OH)2 gave C and D, respectively.

(3) HNO3 oxidation of C and D gave (-)-tartaric acid and meso-tartaric acid, respectively.

(a) You know the absolute configuration of D-(+)-glyceraldehyde, Use Fischer projections to show the absolute configurations of products A, B, C, and D.

(b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and meso-tartaric acid.

(a) Give the products expected when (-)-erythrose reacts with HCN.

(b) What is the relationship between the products? How might they be separated?

(c) Are the products optically active? Explain.

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