EBK ORGANIC CHEMISTRY
8th Edition
ISBN: 8220102744127
Author: Bruice
Publisher: PEARSON
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Chapter 20, Problem 73P
Interpretation Introduction
Interpretation:
The explanation corresponding to the given statement that
Concept Introduction:
The chair conformers are stable forms of cyclic compounds. In chair conformation, substituent’s prefers to place at equatorial positions. This is due to unwanted interactions between those substituent’s that are placed at axial positions. These interactions cause repulsion and that leads to the un-stability in molecule. The general interactions that occur between axial substituents are
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When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react toform an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of d-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 °C, but only about 0.1% of d-glucose exists in the anhydro form under the same conditions.
When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react to form an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of D-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 °C, but only about 0.1% of D-glucose exists in the anhydro form under the same conditions.
The most stable conformation of the pyranose ring of most D-aldohexoses places the largest group, CH2OH, in the equatorial
position. An exception to this is the aldohexose D-idose. Draw the two possible chair conformations of either the a or B anomer
of D-idose. Explain why the more stable conformation has the CH2OH group in the axial position.
Chapter 20 Solutions
EBK ORGANIC CHEMISTRY
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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