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. (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

Erwin Chargaffโ€™s discovery that DNA contains equimolar amounts of guanine and cytosine and also equimolar amounts of adenine and thymine has come to be known as Chargaffโ€™s rule:

G = C and A = T

(a) Does Chargaffโ€™s rule imply that equal amounts of guanine and adenine are present in DNA? That is, does G = A?

(b) Does Chargaffโ€™s rule imply that the sum of the purine residues equals the sum of the pyrimidine residues? That is, does A + G = C + T?

(c) Does Chargaffโ€™s rule apply only to double-stranded DNA, or would it also apply to each individual strand if the double helical strand were separated into its two complementary strands?

Is gentiobiose a reducing sugar? Does it mutarotate? Explain your reasoning.

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 ฮฒ -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?

(e) A similar protecting group, called an acetonide, can block reaction at the 2โ€ฒ and 3โ€ฒ oxygens of a ribonucleoside. This protected derivative is formed by the reaction of the nucleoside with acetone under acid catalysis. From this information, draw the protected product formed by the reaction.

In 1891, Emil Fischer determined the structures of glucose and seven other D-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry and symmetry. He received the Nobel Prize for this work in 1902. Fischer has determined that D-glucose is an aldohexose, and he used Ruff degradation to degrade it to (+)-glyceraldehyde. Therefore, the eight D-aldohexose structures shown in Figure 23-3 are the possible structures for glucose.

Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and sue the following results to prove which of these structures represent glucose, mannose, arabinose, and erythrose.

(a)Upon Ruff degradation, glucose and mannose gives the same aldopentose: arabinose.Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are the two possible structures of arabinose?

(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive aldaric acid, meso-tartaric acid. What is the structure of erythrose?

(c) Which of the two possible structures of arabinose is correct? What are the possible structures of glucose and mannose?

(d) Fischerโ€™s genius was needed to distinguish between glucose and mannose. He developed a series of reactions to convert the aldehyde group of an aldose to an alcohol while converting the terminal alcohol to an aldehyde. In effect, he swapped the functional groups on the ends. When he interchanged the functional groups on D-mannose, he was astonished to find that the product was still D-mannose. Show how this information completes the proof of the mannose structure, and show how it implies the correct glucose structure.

(e) When Fischer interchanged the functional groups on D-glucose, the product was an unnatural L sugar. Show which unnatural sugar he must have formed, and show how it completes the proof of the glucose structure.

After a series of Kilianiโ€“Fischer syntheses on (+)-glyceraldehyde, an unknown sugar is isolated from the reaction mixture. The following experimental information is obtained:

(1) Molecular formula C6H12O6

(2) Undergoes mutarotation.

(3) Reacts with bromine water to give an aldonic acid.

(4) Reacts with HNO3 to give an optically active aldaric acid.

(5) Ruff degradation followed by HNO3 oxidation gives an optically inactive aldaric acid. (6) Two Ruff degradations followed by HNO3 oxidation give meso-tartaric acid.

(7) When the original sugar is treated with CH3I and Ag2O, a pentamethyl derivative is formed. Hydrolysis gives a tetramethyl derivative with a free hydroxy group on C5.

(a) Draw a Fischer projection for the open-chain form of this unknown sugar. Use Figure 23-3 to name the sugar.

(b) Draw the most stable conform

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