Treatment of an organic compound with hot, concentrated hydriodic acid(also known as the Zeisel analysis) yields CH3I, indicating the presence of methoxyl group (-OCH3). Unknown organic compound F (one of the opium alkaloids used in antispasmodic medicine whose formula is C19H21NO3) was treated with hot concentrated hydroidic acid to determine the number of methoxyl groups present in the molecule. When 7.12mg of compound F is treated with hydriodic acid and the CH3I thus formed is passed into alcoholic silver nitrate, 10.73mg of silver iodide is obtained. How many methoxyl groups per molecule of compound F would be indicated by the results of this Zeisel analysis?
Treatment of an organic compound with hot, concentrated hydriodic acid(also known as the Zeisel analysis) yields CH3I, indicating the presence of methoxyl group (-OCH3). Unknown organic compound F (one of the opium alkaloids used in antispasmodic medicine whose formula is C19H21NO3) was treated with hot concentrated hydroidic acid to determine the number of methoxyl groups present in the molecule. When 7.12mg of compound F is treated with hydriodic acid and the CH3I thus formed is passed into alcoholic silver nitrate, 10.73mg of silver iodide is obtained. How many methoxyl groups per molecule of compound F would be indicated by the results of this Zeisel analysis?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Treatment of an organic compound with hot, concentrated hydriodic acid(also known as the Zeisel analysis) yields CH3I, indicating the presence of methoxyl group (-OCH3). Unknown organic compound F (one of the opium alkaloids used in antispasmodic medicine whose formula is C19H21NO3) was treated with hot concentrated hydroidic acid to determine the number of methoxyl groups present in the molecule. When 7.12mg of compound F is treated with hydriodic acid and the CH3I thus formed is passed into alcoholic silver nitrate, 10.73mg of silver iodide is obtained. How many methoxyl groups per molecule of compound F would be indicated by the results of this Zeisel analysis?
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