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: o o a a P A BA A AP A P APA A A P A B A A. A.. .. a a A.. - 1 - [1] - - [2]: ()-N- --[ - ] - CH CH2 CH2 NH CH3 * HCl F254 Merk. O 1 1 CF3 1 [ ] - 1 - 1 1 [ ] - [5-7]. - - Rf - Rf - - Rf=0,04..

- III - 1- - - -1- f I); 1- - - - - II); 1- -1-[ - - ]- - III - - -.

- I, II III - - - Rf=0,2, Rf= Rf= - - Rf F254 Merk - - Rf= -. 1. P. 1. 1 Rf = 0,5; Rf = 0,7; 1 I ; Rf = 0,2; II ; Rf =0,3; III ; Rf = 0, I, II III.

a a ) [8].

, tr 1 P.. Nova-Pak C8 1 - 1 1 - 1 11 1 - - R15025a05m R1a0m 1 X ==, a15R050501,118 R0a11, - R1 Rf, - R0 1 P.. 1 1 0 m 1 11 - C17H18F3NO - = 0,982 + 1,26 10- 1 r = 0,0000, S02 = 4, 1 10-8, b = 0,982, b = 9,8 10-3, a = 1,26 10-4, a = 1, 1 - 10-4 F 1 - 1 1 a 1 1 - P a a B a a a a a a 1 a a Coepae yocea 120 opaa Haeo, % m, r m/m, % Bo m, Haeo, 23 4 5 0,0276 0,0272 -0,0004 -1,4 98,0,0197 0,0195 -0,0002 -1,0 99,0,00985 0,00989 0,00004 0,4 100,0,00394 0,00399 0,00005 1,3 101,0,00197 0,00200 0,00003 1,5 101, P a a a a a a a B Haeo Hoep yocea cep oo acye, 10898 0, 20898 0, 30998 0,B B 41298 0,51298 0, 1 PA P X = 100,2; S = 1,32; P = 95 %; t(95,4) = 2,78;

X = 3,7;

= 1,6; X X = 100/ = 3,7 %; = 100/ = 1,6 %.

1. - 1 1-1 - 2003,. 10.

1 - 3. 1 1 - 4. -. - 1- - 5..11 1 1 1 - 6. - XI 1 1 7. 8.Fedorova G.A., Baram G.I. eta al. Chromatographia, - 2001, v. 53, h.495-497.




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