Organic Chemistry
7th Edition
ISBN: 9780321803221
Author: Paula Y. Bruice
Publisher: Prentice Hall
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Chapter 21, Problem 62P
Interpretation Introduction
Interpretation:
Whether
Concept Introduction:
Carbohydrates are made from carbon, oxygen and hydrogen atoms and the wide numbers of isomers exist for the carbohydrates because of the presence of chiral Carbon atoms.
Pyranose are the compounds that comprise six-membered ring made of one Oxygen and five Carbon atoms, whereas furanose is the carbohydrates that comprise five-membered ring made of one Oxygen and four carbon atoms.
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Students have asked these similar questions
Lactose is a disaccharide in which a glycosidic linkage
connects the monosaccharides galactose and glucose.
OH
НО
OH
(a) Identify the glycosidic linkage and the acetal carbon
in lactose.
(b) What type of glycosidic linkage does lactose have
(i.e., is it 1,1'-, 1,2'-, etc., and is it a or B)?
(c) People who are lactose intolerant are deficient in
the enzyme lactase, and therefore cannot efficiently
break down the disaccharide into its monosaccharides. When lactose is treated with aqueous acid, however, this
hydrolysis can take place, though relatively slowly. Draw the complete, detailed mechanism and the products of the
acid-catalyzed hydrolysis of lactose.
Но
ОН
НО
ОН
ОН
Lactose
Draw the structures (using chair conformations of pyranoses) of the following disaccharides.(a) 4-O-(a-d-glucopyranosyl)-d-galactopyranose(
Trehalose and maltose are both dimers of glucose. However, they have considereably
different reactivities. Concisely explain why these differences are observed.
но
но
HO
HO
но
"HO
он
но
он
OH
O HO
OHOH
но
trehalose
maltose
1. Malthose is a reducing sugar while trehalose is not.
2. Trehalose is very resistant to acid hydrolysis while maltose can be acid-hydrolyzed with
ease.
Chapter 21 Solutions
Organic Chemistry
Ch. 21.1 - Prob. 1PCh. 21.2 - Prob. 2PCh. 21.2 - Prob. 3PCh. 21.3 - Prob. 4PCh. 21.3 - Prob. 5PCh. 21.3 - Prob. 6PCh. 21.4 - Prob. 7PCh. 21.4 - Prob. 8PCh. 21.5 - Prob. 9PCh. 21.5 - Prob. 10P
Ch. 21.5 - Prob. 11PCh. 21.6 - Prob. 12PCh. 21.6 - Prob. 13PCh. 21.6 - Prob. 14PCh. 21.7 - Prob. 15PCh. 21.8 - Prob. 16PCh. 21.9 - Prob. 18PCh. 21.10 - Prob. 20PCh. 21.10 - Prob. 21PCh. 21.10 - Prob. 22PCh. 21.11 - Prob. 24PCh. 21.11 - Prob. 25PCh. 21.15 - Prob. 27PCh. 21.16 - Prob. 28PCh. 21.17 - Prob. 29PCh. 21.18 - Refer to Figure 20.5 to answer the following...Ch. 21 - Prob. 31PCh. 21 - Prob. 32PCh. 21 - Prob. 33PCh. 21 - Prob. 34PCh. 21 - Prob. 35PCh. 21 - Prob. 36PCh. 21 - Prob. 37PCh. 21 - Prob. 38PCh. 21 - Prob. 39PCh. 21 - Prob. 40PCh. 21 - Prob. 41PCh. 21 - Prob. 42PCh. 21 - Prob. 43PCh. 21 - Prob. 44PCh. 21 - The 1H NMR spectrum of D-glucose in D2O exhibits...Ch. 21 - Prob. 46PCh. 21 - Prob. 47PCh. 21 - Prob. 48PCh. 21 - Prob. 49PCh. 21 - Prob. 50PCh. 21 - Prob. 51PCh. 21 - Prob. 52PCh. 21 - Prob. 53PCh. 21 - Prob. 54PCh. 21 - Prob. 55PCh. 21 - A hexose is obtained when the residue of a shrub...Ch. 21 - Prob. 57PCh. 21 - Prob. 58PCh. 21 - Prob. 59PCh. 21 - Prob. 60PCh. 21 - Prob. 61PCh. 21 - Prob. 62PCh. 21 - Prob. 63PCh. 21 - Prob. 64PCh. 21 - Prob. 65PCh. 21 - Prob. 66P
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