Chemical examination of the bark of euonymus atropurpureus / by Harold Rogerson.

  • Rogerson, Harold.
Date:
[1912?]
Catalogue details

Licence: Public Domain Mark

Credit: Chemical examination of the bark of euonymus atropurpureus / by Harold Rogerson. Source: Wellcome Collection.

    11/18 (page 1048)
    A determination of the specific rotatory power gave the following result: 0*2362,*' made up to 20 c.c. with chloroform, gave ccD —40' in a 2-dcm. tube, whence [a]D — 28*2°. The above-described substance gave with acetic anhydride and sulphuric acid a colour reaction similar to that produced by the phytosterols, but differs from the latter, inasmuch as it becomes red when moistened writh strong sulphuric acid. It also differs from the phytosterols by the presence of two oxygen atoms, although containing but one hydroxyl group. As it is a new compound, it is proposed to designate it euonysterol. Euonysteryl Acetate, C31H5102*C0*CH3.—This derivative, when recrystallised from ethyl acetate, separated in laminae melting at 116—118°: 0*1424 gave 0*4156 C02 and 0*1426 H20. C = 79*5; H = ll*l. 0*1174 „ 0*3410 C02 „ 0*1158 H20. C = 79*2 ; H = 10*9. C33H^03 requires C = 79*5; H = 10*8 per cent. Isolation of a New Monohydric Alcohol, Homo-euonysterol, C40H69O-OH. Although euonysterol represented the chief portion of the previ¬ ously described mixture, the more soluble acetyl derivative, when decomposed with alcoholic potassium hydroxide, gave a product which separated from a mixture of ethyl acetate and dilute alcohol in laminae, melting at 133—134°: 0*0997, on heating at 110°, lost 0*0050 H20. H2O = 5*0. 0*0947 * gave 0*2850 C02 and 0*1052 H20. C = 82*l; H = 12*3. C40H70O2,l^H2O requires H20 = 4*6 per cent. C40H70O2 requires C = 82*5; H = 12*0 „ „ This compound is evidently a homologue of the previously described alcohol, differing from it by the increment 9CH2, and it is therefore proposed to designate it homo-euonysterol. Like euony¬ sterol, it became red when moistened with sulphuric acid. Homo-euonysteryl Acetate, C40H69O*CO*CH3.—This derivative separated from acetic anhydride in leaflets melting at 128—130°: 0*1040 gave 0*3068 C02 and 0*1064 H20. C = 80*5; H = ll*4. C42H7203 requires C = 80'7; H = 11'5 per cent. Isolation of a New Dihydric Alcohol, Atropurol, C27H44(OH)2. The original alcoholic mother liquor from the above-described mixture of alcohols was concentrated to about 100 c.c., when, on keeping for a few days, a quantity (2*0 grams) of a crystalline product separated. This was dissolved in a mixture of ethyl * Anhydrous substance.