Sunday, December 13, 2009
Chromatographic Separation Of Purin Bases
Chromatography is an analytical method and technique of separation in which the ingr. Are separated on basis of their difference in migration in a system of two phases. One is fixed on a stationary phase and other is a mobile phase. The difference in migration rates of ingr. Are based on different adsorption partition on exchange molecular sieving effect chief subst. to be separated.
The stationary phase is solid absorbent or a lia. Feed on a solid carrier. Mobile phase travel through stationary phases it carries with it ingr. From sample at different rates. The technique is widely used for separation of amino acids, lipids, alkaloids etc. TLC of Purina bases viz. adenine and Uralic stationary phase: silica gel supported as glass plate.
Mobile phase: - n-butane, n-methanol, H2O, NH3 in ratio 60:20:2:1.
a) Purina bases show weak flu ore séance in U.V.light.
b) Spray of 0.2% cosign in ethanol sate. C mercuric chloride which forms a complex with bases is appearing as black spot in U.V.light.
Procedure:- Preparation of plate: Prepare 10% slurry of silica gel in H2O and spread it evenly on a glass plate and allow it to set – dry it at 1100c and wol.
Soln of chamber: - Preparation a solvent ys. And pour about 20 ml in a TLC chamber and put a pica of filter paper in it close in air tight container keep it from nearly 30 mins.
Preparation of sample: - Prepare 1% w/w soln of adenine and Uralic in hot water and sy of them.
Spotting: Spot about 0.05 ml of each sample and each std. soln on TLC and dry in air at r.t.note that spots don’t spread two much are sufficient apart.
Developing: - Pat the plate. In previously sold chamber and allow to develop till the solvent runs to a distance of 20 cms. Mark the solvent front, take out the plates and keep as flat sur. Dry in air.
Location: - Observe the plates as such under UV or spray the plate c reagent. Allow to stand for 15 min and observe under UV the spots and calculate RP value.
Observation:-
Dist. Traveled by A = 10 cms.
Dist. Traveled by B = 8 cms.
Dist. Traveled by solvent = 15 cms.
Rf = Dist.travelled by samples .
Dist.travelled by solvent front.
Therefore Rf of A = 10/15 = 0.66
Rf of B = 5 /15= 0.53
Result:-
Rf of A = 0.66
Rf of B = 0.53
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