Organic Chemistry, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (8th Edition)
8th Edition
ISBN: 9780134473147
Author: Paula Yurkanis Bruice
Publisher: PEARSON
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Question
Chapter 20, Problem 51P
Interpretation Introduction
Interpretation:
Compounds A, B, C and D are to be identified.
Concept Introduction:
The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. A sugar molecule in which hydroxyl groups of second last carbon atom is on left side is known as
The Wohl degradation is used in order to reduce the chain of an aldose by one carbon and Kiliani-Fischer synthesis is used to increase the chain length of an aldose sugar.
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An unknown reducing disaccharide is found to be unaffected by invertase enzymes. Treatment with an a@galactosidasecleaves the disaccharide to give one molecule of d-fructose and one molecule of d-galactose. When the disaccharideis treated with excess iodomethane and silver oxide and then hydrolyzed in dilute acid, the products are2,3,4,6-tetra-O-methylgalactose and 1,3,4-tri-O-methylfructose. Propose a structure for this disaccharide, and give itscomplete systematic name.
A D-aldopentose A is reduced to an optically active alditol. Upon Kiliani–Fischer synthesis, A is converted to two Daldohexoses, B and C. B is oxidized to an optically inactive aldaric acid. C is oxidized to an optically active aldaric acid. What are the structures of A–C?
A D-aldohexose A is formed from an aldopentose B by the Kiliani–Fischer synthesis. Reduction of A with NaBH4 forms an optically inactive alditol. Oxidation of B forms an optically active aldaric acid. What are the structures of A and B?
Chapter 20 Solutions
Organic Chemistry, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (8th Edition)
Ch. 20.1 - Prob. 1PCh. 20.2 - Prob. 2PCh. 20.2 - Prob. 3PCh. 20.3 - Prob. 4PCh. 20.3 - Prob. 5PCh. 20.3 - Prob. 6PCh. 20.4 - Prob. 7PCh. 20.4 - Prob. 8PCh. 20.5 - Prob. 9PCh. 20.5 - Prob. 10P
Ch. 20.5 - Prob. 11PCh. 20.6 - Prob. 12PCh. 20.6 - Prob. 13PCh. 20.6 - Prob. 14PCh. 20.7 - Prob. 15PCh. 20.8 - Prob. 16PCh. 20.9 - Prob. 18PCh. 20.10 - Prob. 20PCh. 20.10 - Prob. 21PCh. 20.10 - Prob. 22PCh. 20.11 - Prob. 23PCh. 20.11 - Prob. 24PCh. 20.12 - Prob. 25PCh. 20.12 - Prob. 26PCh. 20.14 - Prob. 28PCh. 20.15 - Prob. 29PCh. 20.15 - Prob. 30PCh. 20.16 - Prob. 31PCh. 20.17 - Prob. 32PCh. 20.18 - Refer to Figure 20.5 to answer the following...Ch. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - The 1H NMR spectrum of D-glucose in D2O exhibits...Ch. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - A hexose is obtained when the residue of a shrub...Ch. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 67PCh. 20 - Prob. 68PCh. 20 - Prob. 69PCh. 20 - Prob. 70PCh. 20 - Prob. 71PCh. 20 - Prob. 72PCh. 20 - Prob. 73P
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