Chemical examination of Ipomoea purpurea / by Frederick B. Power and Harold Rogerson.

  • Frederick Belding Power
Date:
[1908?]
Catalogue details

Licence: In copyright

Credit: Chemical examination of Ipomoea purpurea / by Frederick B. Power and Harold Rogerson. Source: Wellcome Collection.

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    of potassium hydroxide, the alcohol then removed, and the residual thick, green liquid poured into a large volume of water. The strongly alkaline liquid was extracted six times with ether, the com¬ bined ethereal solutions washed with water, dried with calcium chloride, and the ether removed, when a dark-colored liquid was obtained which, on cooling, solidified to a crystalline mass. This was dissolved in alcohol, the solution heated with animal charcoal, and filtered, when, on cooling, a quantity of a solid substance sepa¬ rated, which was collected on a filter and spread on a porous plate. After being again subjected to the same process of purification, it was finally distilled under a pressure of 15 m.m , ween it was obtained as a white, silky mass. This was crystallized from ethyl acetate, from which it separated in small, lustrous leaflets, melting sharply at 74-75° C. 0-0794 gave 0-2480 C0.2 and 0-1042 H20. C — 85-2 ; H = 14 6 C35H72 requires C = 85-4; H = 14.6 per cent. The above described substance was thus identified as pentatria- contane. The alcoholic mother-liquors from the first crystallization of the pentatriacontane were concentrated and allowed to stand, when a further portion of a solid substance separated. This was collected, dried on a porous plate, and subsequently crystallized several times from a mixture of ethyl acetate and alcohol containing a little water. It was then obtained in colorless laminae, melting at 132-1330 C., and afforded the color reactions characteristic of the phytosterols. On analysis it gave the following results: 0-1912 of the air-dried substance when heated to 105° C. lost 0-0090. H20 — 4-7 0 0682 of anhydrous substance gave 0-2094 CCh and 0-0722 H20 C = 837 ; H = 1 r8 C.^H^ O, H20 requires H20 = 4-5 per cent. C^H^O requires C = 83 9 ; H = 11*9 per cent. This substance was thus identified as a phytosterol. Its optical rotatory power was determined with the following result:— 0-2336 of anhydrous substance, dissolved in 25 c.c. of chloroform, gave aD — 0-36° in a 2 dcm. tube, whence [a]D — 3210. From the character of this phytosterol it would appear probable