Out Line Phyt
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No Group tests Procedure Sources
1
AkaloidsDragendoff’s test 0.2 g sample diluted in 2% H2SO4 warmed (2 m) filtered filtrate +
Dragendoff’s reagent orange red precipitate (+)Uddin, et al., 2011
1 ml sample solution + 2 ml Dragendoff’s reagent orange red precipitate (+) Madppa and Bapaih, 2012Prepare reagentSolution A : Bismuth nitrate (0.17g) in AcOH (2 mL) and H2O (8 mL), solution B: KI (4 g) in AcOH (10 mL) and H2O (20 mL) Mix Solutions A and B and dilute to 100 mL with H2OWe need 12 ml (2 sample x 3 replications x 2 ml = 12 ml) ~ 25 mlBismuth nitrate = 0.17 g / 4 = 0.425 gAcetic acid = 12 ml / 4 = 3 mlPotassium iodide = 4 g / 4 = 1 gram
Mayer’s test Sample solution + Mayer’s reagent (Potassium Mercuric Iodide) yellow colored precipitate (+)
Tiwari, et al., 2011
Prepare reagentcreated by dissolving 1.358 grams of HgCl2 in 60 ml of water, and pouring that solution into a solution of 5 grams of KI in 10 ml of waterWe need 25 mlMercuric(II) chloride = (1.358 g / 80 ml) x 25 ml = 0.485 g Potassium iodide = (5 g / 80 ml) x 25 ml = 1.5625 ~ 2 g
Wagner’s test Sample solution + Wagner’s reagent reddish brown precipitate/coloration (+) Tiwari, et al., 2011Sample solution + 3-5 drops of Wagner’s reagent [1.27g of iodine and 2g of potassium iodide in100 ml of water] reddish brown precipitate/coloration (+)
Ugochukwu, et al., 2013
Prepare reagent1.27g of iodine and 2g of potassium iodide in100 ml of waterWe need 10 mlIodine = 1.27 g / 10 = 0.127 gPotassium iodide = 2 g / 10 = 0.2 g
2 Anthraquinones 0.5 g sample boiled in 10% HCl (few m in water bath) filtered cooled filtrate + CHCl3 (equal volume) + few drops 10% NH3 heated rose pink (+)
Uddin, et al., 2011
1 ml sample + 10 ml benzene filtered filtrate + 5 ml 10% NH3 shaken pinkish (+)
Madppa and Bapaih, 2012
Prepare reagentBenzene = 2 sample x 3 replications x 10 ml = 60 ml Conc. NH3 (25%) = (30 ml x 10%) / 25% = 12 ml, volume of 10% NH3 = 30 ml (2 sample x 3 replications x 5 ml = 30 ml)
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Antocyanin 1 ml sample solution + 5 ml diluted HCl pink colour (+) Madppa and Bapaih, 2012Prepare reagent1% HCl, 2 sample x 3 replications x 5 ml = 30 ml ~ 50 mlConc. HCl (37%) = (50 ml x 1%) / 37% = 1.35 ml
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TanninsFeCl3 test Sample (a small quantity) diluted in H2O heated in water bath filtered filtrate
+ few drops FeCl3 dark green solution (+)Uddin, et al., 2011
1 ml sample solution + 2 ml FeCl3 dark green solution (+) Madppa and Bapaih, 2012Prepare reagent 5% FeCl3, 2 sample x 3 replications x 2 ml = 12 ml ~ 25 mlIron(III) chloride = (5/100) x 25 ml = 1.25 gIron(III) chloride hexahydrate = (1.25 g x 270.3) / 162.2 = 2.08 g
Gelatin test Sample solution + gelatin solution white precipitate (+) Bhandary, et al., 2012Sample solution + 1% gelatin containing NaCl white precipitate (+) Tiwari, et al., 2011Prepare reagent1% gelatin, 25 mlGelatin = (1/100) x 25 ml = 0.25 gSodium chloride = a few ~ 1 g
Vanilin- hydrochloric acid test
Sample solution + vanillin-hydrochloric acid reagent a pink or red color (+)
Devmurari, 2010
Prepare reagentVanillin 1 g, alcohol 10 mL, concentrated hydrochloric acid 10 mLVanillin = 1 gEthanol = 10 mlConc. HCl = 10 ml
5 Cadiac glycosidesLieberman buchard test Journal in Thai languageKeller Kelliani’s test 5 ml sample solution + 2 ml Acetic glacial + a drop FeCl3 + 1 ml conc. H2SO4 (addition
acid should be carefully to keep the mixture in 2 layers) brown ring at the interface indicated the presence of deoxysugar characteristic of cardenolides. A violet ring may
Ugochukwu, et al., 2013
appear below the ring while in the acetic acid layer, a greenish ring may formPrepare reagentAcetic glacial = 2 sample x 3 replications x 2 ml = 12 mlConc. Sulphuric acid = 2 sample x 3 replications x 1 ml = 6 mlFeCl3, use tannins – FeCl3 test
Modified Borntragers test Sample solution + diluted HCl heated + FeCl3 immersed in boiling water for 5 m + benzene (equal volume) benzene layer + NH3 solution rose pink (+)
Tiwari, et al., 2011
Prepare reagentConc. HCl (37%), use antocyanin testFeCl3, use tannins – FeCl3 testBenzene = 2 sample x 3 replications x 2 ml = 12 ml
FeCl3 reagent 1 ml sample solution + FeCl3 reagent (5% FeCl3 : Ac. Glacial = 1:99) + Few drops conc. H2SO4 greenish blue colour (+)
Madppa and Bapaih, 2012
Prepare reagentIron(III) chloride = (5/100) x 1 ml = 0.05 gAcetic glacial = 99 ml
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Saponins (foam test) Sample solution + H2O shaken stable froth for 15 m (+) Bhandary, et al., 20121 ml sample solution + 2 ml H2O shaken allowed 10 m eternal foam (+) Madppa and Bapaih, 20122 ml sample solution + 6 ml H2O shaken persistent foam (+) Ugochukwu, et al., 20130.2 mg sample in 5 ml H2O shaken boiled Frothing (+) Uddin, et al., 2011
7 FlavonoidsAlkaline reagent test Sample solution + few drops diluted NaOH yellow mixture + few drops diluted
HCl colorless (+)Uddin, et al., 2011 and Bhandary, et al., 2012
2 ml sample solution + few drops 20% NaOH yellow mixture + diluted HCl colorless (+)
Ugochukwu, et al., 2013
1 ml sample solution + 2 ml diluted NaOH golden yellow (+) Madppa and Bapaih, 2012sample solution + few drops NaOH solution yellow + HCl solution colorless (+)
Khanam, et al. 2013
Prepare sampleSodium hydroxide = (20/100) x 10 ml = 2 g, Volume of 20% NaOH = 10 mlConc. HCl (37%), use antocyanin test
FeCl3 test Sample solution + few drops FeCl3 blackish red (+) Bhandary, et al., 2012Prepare reagentFeCl3, use tannins – FeCl3 test
Pb-Ac test Sample solution + Pb-Acetate yellow precipitate (+) Tiwari, et al., 2011Sample solution + few drops 10% Pb-Acetate yellow precipitate (+) Bhandary, et al., 2012Prepare sampleLead acetate = (10/100) x 10 ml = 1 g, Volume of 10% Pb-Acetate = 10 ml
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SteroidsLieberman Burchard test Sample + acetic anhydride boiled cooled mixture + conc. H2SO4 (addition acid
should be carefully to keep the mixture in 2 layers) brown ring at the junction of two layers, green coloration of the upper layer and the formation of deep red color in the lower layer (+)
Bhandary, et al., 2012
0.5 g sample + 2ml H2SO4 + 2 ml acetic anhydride violet to blue/green (+) Uddin, et al., 2011Prepare reagentAcetic anhydride = (2 sample x 3 replications x 1 ml) = 6 mlConc. Sulphuric acid = few ~ 3 ml
Salkowski test Mixture of 1 ml sample solution and 10 ml CHCl3 + 10 ml conc. H2SO4 (addition acid should be carefully to keep the mixture in 2 layers) upper layer, red and H2SO4 layer, yellowish green (+)
Madppa and Bapaih, 2012
Prepare reagentLook terpenoids test
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PhenolPhenol-flavonoids (Pb-Ac) 1 ml sample solution + 2 ml 10 % Pb-Acetate brown precipitate (+) Madppa and Bapaih, 2012
Prepare reagentLook Flavonoids test
FeCl3 test Sample solution + 5% FeCl3 blue or black color (+) Ugochukwu, et al., 2013Sample solution + FeCl3 bluish black color (+) Tiwari, et al., 20115 ml of 10 mg/ml sample solution + few drops 5% FeCl3 Bluish black color (+) Khanam, et al. 2013Prepare reagentFeCl3, use tannins – FeCl3 test
Folin Ciocalteu10 Terpenoids
(Salkowski test)0.2 g sample diluted in 2 ml CHCl3 mixture + 3 ml conc. H2SO4 (addition acid should be carefully to keep the mixture in 2 layers) reddish brown at interface (+)
Uddin, et al., 2011
1 ml sample solution + 3 ml CHCl3 mixture + few drops conc. H2SO4 (addition acid should be carefully to keep the mixture in 2 layers) red colour (+)
Madppa and Bapaih, 2012
2 ml sample solution + 1 ml CHCl3 + few drops conc. H2SO4 dark pink/red color (+) Ugochukwu, et al., 2013
0.5 mg extract + few ml of CHCl3 + few ml of conc. H2SO4 (addition acid should be Khanam, et al. 2013
carefully to keep the mixture in 2 layers) Reddish brown color at interface (+)Prepare reagentChloroform = 2 samples x 3 replications x 2 ml = 12 mlConc. Shulpuric acid = 2 samples x 3 replications x 3 ml = 18 ml