9. Treatment of an organic compound with hot, concentrated hydriodic acid (also
ID: 1075972 • Letter: 9
Question
9. Treatment of an organic compound with hot, concentrated hydriodic acid (also known as the Zeisel analysis) yields CHyI, indicating the presence of a methoxyl group (-OCHs). Unknown organic compound F (one of the opium alkaloids used in antispasmodic medicine, whose formula is CioH21NO) was treated with hot concentrated hydriodic acid to determine the number of methoxyl groups present in the molecule. When 7.12 mg of compound F is treated with hydriodic acid and the CHsl thus formed is passed into alcoholic silver nitrate, 10.73 mg of silver iodide is obtained. How many methoxyl groups per molecule of compound F would be indicted by the results of this Zeisel analysis? (2 points)Explanation / Answer
The reaction between CH3I and AgNO3 will be
CH3I + AgNO3 ------ CH3NO3 + AgI
Molar mass of silver iodide (AgI) = 234.77 gm/mol
Number of moles of silver iodide = mass/molar mass = 10.73 * 10^(-3)/234.77 = 4.5704 * 10^(-5) moles
Since 1 mole of AgI is produced from 1 mole of CH3I, hence number of moles of CH3I = 4.5704 * 10^(-5) moles
Molar mass of F (compound - C19H21NO3) = 19 * 12 + 21 * 1 + 14 + 3 * 16 = 311 gm/mol
Numbe of moles of F = Mass/molar mass = 7.12 * 10^(-3)/311 = 2.289 * 10^(-5) moles
Number of methoxy groups in compound = Number of moles of CH3I/Number of moles of F = 4.5704 * 10^(-5) moles/ 2.289 * 10^(-5) moles = 2
Hence compound F contains two methoxy groups
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