Friday, February 8, 2013

Warfarin

Warfarin is an anticoagulant also known as Coumadin. It is an odorless, colorless or white crystalline powder available commercially as a dust or liquid concentrate in various formulations. It is also found in anticoagulant medications and pesticides.
·         IUPAC name: (RS)-4-hydroxy- 3-(3- oxo- 1-phenylbutyl)- 2H- chromen- 2-one

·         2-dimensional structure:
 

·         3-D Structure:


·         Chemical formula: C19H16O4

·         Molecular weight: 308.3 g/mol

·         Physical properties

o    Melting Point: 161⁰C
    • Boiling points: Decomposes at 760 torr
    • H2O solubility: Insoluble in H2O
    • Benzene Solubility: Insoluble in Benzene

Intermolecular forces present in the molecule: Warfarin has dispersion forces, as well as dipole- dipole forces. It is capable of forming hydrogen bonds with other molecules. Warfarin also has 41 sigma bonds and 9 pi bonds.

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6 comments:

  1. Determine the vapor pressure (in mm Hg) of substance A at 5 degrees Celsius, given that it's normal boiling point is 76 degrees Celsius and that the ΔHvap is 38.7 kJ/mol.

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  2. Replies
    1. What is this? No calculations or any work is shown?

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  3. To solve this problem, you must use the Clausius Clapyron equation.
    ln (P2/P1)= (-ΔHvap/R)*(1/T2-1/T1)
    We can use this form of the equation to predict the vapor pressure of a liquid at any temperature if we know ΔHvap and the normal boiling point. R is a constant in this equation (8.314J/mol*K). We must also be sure to convert the given temperatures from Celcius to Kelvin and convert pressure in kJ to Joules. 1 torr= 1mmHg.We assume that P1 is equvalent to 1 atm which is equivalent to 760mmHg.

    ln (p2/p1)= (-38700J/mol)/(8.314J/K*mol)*((1/278K)-(1/349K))
    =-3.41
    p2/p1= e^-3.41
    P2=p1*e^(-3.41)
    P2=(760mmHg)(.033)
    P2=25.11mmHg

    Source: Chemistry: A Molecular Approach, Tro

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