Saponarin : a new glucoside, coloured blue with iodine / by George Barger.

  • George Barger
Date:
[1906?]
Catalogue details

Licence: Public Domain Mark

Credit: Saponarin : a new glucoside, coloured blue with iodine / by George Barger. Source: Wellcome Collection.

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    The molecular weight was determined* in pyridine solution (a) with the substance dried at 100° : 0 299 gram in 2*96 grams of pyridine was between 0'223 and 0*238 mol,, benzil as standard, ordinary tempera¬ ture, hence M = 424—453, mean 438; (b) with the substance dried in a vacuum, the pyridine dried over caustic potash : 0*209 gram in 1*98 grams pyridine was between 0*22 and 0*23 mol., benzil as standard, temperature 90°, hence M = 459 —480, mean 469. C21H24012 requires 468. The lead salt was also analysed. It was prepared by precipitating a solution of the glucoside in dilute ammonia with neutral lead acetate, filtering, and washing the yellow precipitate by suspension in hot water. The lead salt was dried in a vacuum until constant in weight and completely decomposed by heating with fuming nitric acid in a sealed tube, the lead being then estimated as sulphate. 0 3489 gave 0*2444 PbS04. Pb = 47*8. C2lHon012Pb.2 requires Pb = 47*2 per cent. Ennea-acetylsaponarin. The glucoside is readily acetylated by boiling for a few seconds with a large excess of acetic anhydride and a few drops of concentrated sulphuric acid. The acetyl derivative is easily soluble in hot alcohol and crystallises on cooling in microscopic, curved needles, melting at 183—185°. It is extremely soluble in ethyl acetate, benzene, and chloroform, and was purified by dissolving in ethyl acetate, which leaves a small quantity of an impurity behind, and then crystallising from a dilute solution in alcohol; 0*915 gram in 10 c.c. of ethyl acetate gave, in a 1-dem. tube, aD - 4*90°, whence [a]D -5*33°. 0*1436 gave 0*2911 C02 and 0*0618 H20. C = 55*29 ; H = 4 78. 0*2078 gram in 0*8586 gram of ethyl acetate was intermediate between 0*2875 and 0*300 mol. benzil. Hence M = 807—842, mean = 825. C21 Hi5O12(C0H3O)9 requires C = 55*32 ; H = 4*96 percent. M = 846. The number of acetyl groups was determined by A. G. Perkin’s ethyl acetate method (Trans., 1905,87, 107). 1. 0*4246 gram gave 0*2748 gram acetic acid = 64*8. 2. 0*7223 „ „ 0*4632 „ „ „ = 64*1. Theory for nine acetyl groups requires 63*8 per cent. * All molecular weight determinations in this paper were made by the author’s microscopic method (Trans., 1904, 85, 286; 1905, 87, 1750).
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