Physical Properties of C21H34O2:
-------------------------------

Boiling point = 430 - 470°C; the only way this substance would get airborne is via fire, no one could physically vaporize this with conventional means of conduction and radiation.
Melting point = Oils are solid via hydrogenation or freezing, but no one has frozen urushiol yet. It is possible to freeze oils (like a block of ice).
SUBSTANCE     FREEZING POINT
corn oil       - 20°C       Table Courtesy of:
sunflower oil  - 17°C       http://www.newton.dep.anl.gov/askasci/chem03/chem03265.htm
olive oil      - 6°C
sesame oil     - 6°C        Looking at this table, one could say the urushiol may
peanut oil       3°C        freeze at a low~mid temperature due to the fact the IMFs
palm oil         24.1°C     holding the molecules together are London Dispersion
coconut oil      25.1°C     Forces, the (base) weakest intermolecular forces, that
                            grow in strength as the complexity of the molecules
                            increase.
Density = 0.973 g/cm3 (it's an oil, oils are most likely going to be less dense (floats) on water)

Color: Clear liquid or pale yellowish/turpentine tint



Chemical Properties of C21H34O2:

Solubility = Since this is an oil, the substance is highly soluble in water, due to density differences, but readily soluble in ethers (organic compounds that is two alkyl or aryl groups connected to an Oxygen atom) and alcohols (organic compounds where a carbon atom is connected to a hydroxyl group [-OH]).

Ever see one of those rainbow colored oil pools in the road? That's because of miscibility, the ability for liquids to form homogenous mixtures (solutions). Petroleum is very
miscible with H2O(L) (water) and thus those puddles form. The same is to be said when urushiol reacts with ethers, alcohols, and benzene.
This is the general denoted formula of an ether R - O - R'
This is the general denoted formula of an ether R - O - R'
This is a picture of alcohol
This is a picture of alcohol


Elemental Composition: A cathecol ring with a C
15 side chain
  1. C% = 79.19% ----> 12.010g * 21 =.....252.210 g
  2. O% = 10.05% ----> 16.000g * 2 .......= ......32.000 g
  3. H% = 10.76% ----> ........1.008g........* 34 = . 34.272 g
......................................../.......................................................................................................................................................... .Molar Mass =.....318.482 g

Formation and Acquisition:

  1. This is a naturally occurring allergen found within plants of the Anacardiaceae family
  2. It has not been synthesized in a lab yet, I believe.
  3. This is legal since all you have to do is extract it from a very common plant.
    • There are several ways you can extract pure samples of urushiol. Usually, urushiol is exacted via liquid-liquid "solvent" extraction. In this process, an aqueous or liquid solution is added to an organic solvent or water. The hopes are that certain compounds in the composition of a substance are immiscible with the solvent or water while others are highly miscible. By doing such, the immiscible components can be removed while the miscible components can be further separated by other means. Another possibility is liquid chromatography. As a hydrocarbon, urushiol is very non-polar. It would be one of the first substances to be exuded since in liquid chromatography, the column/stationary phase of the process is polar. That means the non-polar substances will exit first with the moving phase. If it was reverse liquid chromatography, the order would be vice-versa, but ultimately the results would be the same.

Combustion of urushiol:
2
C21H34O2(L)+57O2(g) ---> 34H2O(g)+ 42CO2(g)