Why does the specific rotation of a freshly prepared solution of the a form gradually decrease with time? Why do solutions of the a and ß forms reach the same specific rotation at equilibrium? Glucose interconverts between cyclic and linear structures. The linear structure is most favorable. When the system reaches equilibrium, the net amounts of each structure no longer change. Glucose interconverts between pyranose and furanose forms. The furanose form is most favorable. When the system reaches equilibrium, the net amounts of each form no longer change. Glucose interconverts between a and ß configurations. The ß configuration is the most favorable. When the system reaches equilibrium, the net amounts of each configuration no longer change. Calculate the percentage of each of the two forms of D-glucose present at equilibrium. a-D-glucose: B-D-glucose: % %

Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter31: Immunochemistry
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Problem 31.75P
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A freshly prepared solution of αα ‑D‑glucose shows a specific rotation of +112°.+112°. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [?]25 °CD=+52.5°.[α]D25 °C=+52.5°. In contrast, a freshly prepared solution of ?β‑D‑glucose has a specific rotation of +19°.+19°. The rotation of this solution increases over time to the same equilibrium value as that shown by the ?α anomer. A solution of one enantiomer of a given monosaccharide rotates plane‑polarized light to the left (counterclockwise) and is the levorotatory isomer, designated (−). The other enantiomer rotates plane‑polarized light to the same extent but to the right (clockwise) and is the dextrorotatory isomer, designated (+). An equimolar mixture of the (+) and (−) forms does not rotate plane‑polarized light.

The optical rotation, the number of degrees by which plane‑polarized light rotates on passage through a given path length of a solution of the compound at a given concentration, quantitatively describes the optical activity of a stereoisomer. The specific rotation ([?]??)([α]λt) of an optically active compound is specific for a particular temperature (?)(t) and wavelength of light (?).(λ). Usually, as here, ?λ is the D line of sodium, 589 nm.

O Macmillan Learning
Which Haworth perspective formula represents B-D-glucose?
НО,
1
3
4
2
ОН
НО,
ОН
ОН
ОН
1
3
- O
ОН
ОН
O OH
О
ОН
ОН
НО
Но
НО
Но
НО
2
4
О ОН
ОН
НО
ОН
Transcribed Image Text:O Macmillan Learning Which Haworth perspective formula represents B-D-glucose? НО, 1 3 4 2 ОН НО, ОН ОН ОН 1 3 - O ОН ОН O OH О ОН ОН НО Но НО Но НО 2 4 О ОН ОН НО ОН
Why does the specific rotation of a freshly prepared solution of the a form gradually decrease with time? Why do solutions
of the a and ß forms reach the same specific rotation at equilibrium?
Glucose interconverts between cyclic and linear structures. The linear structure is most favorable. When the system
reaches equilibrium, the net amounts of each structure no longer change.
Glucose interconverts between pyranose and furanose forms. The furanose form is most favorable. When the system
reaches equilibrium, the net amounts of each form no longer change.
Glucose interconverts between a and ß configurations. The ß configuration is the most favorable. When the system
reaches equilibrium, the net amounts of each configuration no longer change.
Calculate the percentage of each of the two forms of D-glucose present at equilibrium.
a-D-glucose:
B-D-glucose:
%
%
Transcribed Image Text:Why does the specific rotation of a freshly prepared solution of the a form gradually decrease with time? Why do solutions of the a and ß forms reach the same specific rotation at equilibrium? Glucose interconverts between cyclic and linear structures. The linear structure is most favorable. When the system reaches equilibrium, the net amounts of each structure no longer change. Glucose interconverts between pyranose and furanose forms. The furanose form is most favorable. When the system reaches equilibrium, the net amounts of each form no longer change. Glucose interconverts between a and ß configurations. The ß configuration is the most favorable. When the system reaches equilibrium, the net amounts of each configuration no longer change. Calculate the percentage of each of the two forms of D-glucose present at equilibrium. a-D-glucose: B-D-glucose: % %
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