Chapter 23: Q21P (page 1225)
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.
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Chapter 23: Q21P (page 1225)
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.



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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.
Which of the following sugars are reducing sugars? Comment on the common name sucrose for table sugar.
(a) (b) (an aldohexose)
(c) (d)
(e) (f)

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) Fischer鈥檚 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.
Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of is +150.70 , and that of the 尾 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.
H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5鈥 OH group of nucleosides.

The trityl ether derivative of just the 5鈥 OH group is obtained by reaction of the nucleoside with trityl chloride, MMT chloride, or DMT chloride and a base like Et3N. The trityl ether derivative can be removed in dilute aqueous acid. DMT derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest.
(a) Draw the product with the trityl derivative on the 5鈥 oxygen.
(b) Explain why the trityl derivative is selective for the 5鈥 OH group. Why doesn鈥檛 it react at 2鈥 or 3鈥? (c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange when acid is added to a DMT derivative?
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