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Draw the structures of the compounds named in Problem 23-20 parts (a), (c), and (d). Allose is the C3 epimer of glucose and ribose is the C2 epimer of arabinose.

Short Answer

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Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

Step by step solution

01

Pyranose and furanose

Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

02

Epimers

Sugars which have the common configuration like the arrangement of H - and - OH around carbon atoms are same except the carbon present at α -position or carbon number 2 are known as epimers. For example, glucose and mannose are epimers of each other.

03

Anomers

The diastereomers resulting from cyclisation are known as anomers. They differ only in the configuration around first carbon (C1) which is referred to as the anomeric carbon (hemiacetal carbon atom).

04

Identifying α and β anomers of pyranoses and furanoses

For α -anomer, hydroxy group (-OH) on the anomeric carbon is down (axial position) while for β -anomer, hydroxy group (-OH) on the anomeric carbon is up (equatorial position). Again, for α -anomer, the anomeric hydroxy (-OH) group is placed trans to the terminal -CH2OH group while for β -anomer, the anomeric hydroxy (-OH) group is placed cis to the terminal -CH2OH group.

05

Structure of the sugars

In chair conformation of methyl α -D-galactopyranoside, the anomeric carbon is C1. The methyl group (-CH3) on the anomeric carbon is down (axial position).

In chair conformation of α -D-galactopyranoside, the anomeric carbon is C1. The hydroxy group (-OH) on the anomeric carbon is down (axial position).

In Haworth projection of β-D-ibofuransoide, the anomeric carbon is C1. The ethyl group (-CH2CH3)on the anomeric carbon is up which is cis to the terminal -CH2CH group.

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Most popular questions from this chapter

Question:

a) Draw D-allose, the C3 epimer of glucose.

b) Draw D-talose, the C2 epimer of D-galactose.

c) Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.

d) Draw the C4 “epimer” of D-xylose. Notice that this “epimer” is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

Which of the sugars mentioned in Problem 23-53 and 23-54 are reducing sugars? Which ones undergo mutarotation?

. (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.

Cellulose is converted to cellulose acetateby 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.

Question. Retroviruses like HIV, the pathogen responsible for AIDS, incorporate an RNA template that is copied into DNA during infection. The reverse transcriptaseenzyme that copies RNA into DNA is relatively nonselective and error-prone, leading to a high mutation rate. Its lack of selectivity is exploited by the anti-HIV drug AZT (3’-azido-2’,3’-dideoxythymidine), which becomes phosphorylated and is incorporated by reverse transcriptase into DNA, where it acts as a chain terminator. Mammalian DNA polymerases are more selective, having a low affinity for AZT, so its toxicity is relatively low.

  1. Draw the structures of AZT and natural deoxythymidine.
  2. Draw the structure of AZT 5’-triphosphate, the derivative that inhibits reverse transcriptase.
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