1.0) Description: Clear Colourless liquid with a distinct lemon flavor and aroma.
2.0) Solubility: Soluble in
- Tests:
3.1) Specific gravity : between 0.849 and 0.855.
3.2) Angular rotation : between +57o and +65.6o .
3.3) Refractive index : between 1.473 and 1.476 at 20o .
3.4) Ultraviolet absorbance:
Transfer about 250 mg of Oil, accurately weighed, to a 100-mL volumetric flask, add alcohol to volume, and mix. Record the UV absorption spectrum of this solution from 260 to 400 nm in a 1-cm cell, using alcohol as the blank. Determine the absorbance at the wavelength of maximum absorbance at about 330 nm using the line drawn tangent to the curves appearing as minima in the spectrum in wavelength regions above and below the maximum wavelength as the baseline. The absorbance, calculated on the basis of a 250-mg specimen, is not less than 0.20 for California-type Orange Oil, or not less than 0.49 for Florida-type Orange Oil.
3.5) Heavy metals : 0.004%.
3.6) Foreign oils:
Place 50 mL of Oil in a four-bulb Ladenburg flask having the following dimensions: the lower or main bulb is about 6 cm in diameter, and the smaller condensing bulbs are about 3.5, 3.0, and 2.5 cm in diameter; the distance from the bottom of the flask to the side-arm is about 20 cm. Distill Oil at a rate of 1 drop per second until the distillate measures 5 ml, the angular rotation of the first 5 ml is not more than 6o less than of the original oil. The refractive index at 20o of this same portion is between 0.001 and 0.003 lower than that of the original oil.
3.7) Assay
Dissolve 4.5 g of hydroxylamine hydrochloride in 13 mL of water, add 85 mL of tertiary butyl alcohol, mix, and adjust with 0.5 N potassium hydroxide to a pH of 3.4. Pipet 50 mL of this solution into a conical flask containing about 5 mL of Oil, accurately weighed. Insert the stopper in the flask, and allow to stand at room temperature for 30 minutes, with occasional shaking. Titrate the liberated hydrochloric acid with 0.5 N alcoholic potassium hydroxide VS to a pH of 3.4. Each mL of 0.5 N alcoholic potassium hydroxide consumed in the titration is equivalent to 78.13 mg, of total aldehydes, calculated as decanal (C10H20O).
