Qualitative analysis 2 Flashcards
Bayer’s test reagent
alkaline/neural KMnO4
MnO4- gets reduced to MnO2
alkene becomes vicinal diol
Cerric ammonium nitrate formula
(NH4)2[Ce(NO3)6]
Cerric ammonium nitrate test equation
(NH4)2[Ce(NO3)6] + ROH ——–> [Ce(NO3)4 (ROH)3] + NH4NO3
Neutral FeCl3 diff colours for different phenolic compounds
phenol: violet
cresol: violet
catechol: green
resorcinol: violet
1 and 2 napthol: no characteristic colour
pthalein dye
phenol + pthalic anhydride (in oresence of H2SO4) gives phenolphthalein
which gives a pink color on addition of NaOH
Phenolphthalein structure
Take pthalic anhydride. Leave one of the CO grps as it is. For the other one, replace the double O of the C=O with two phenols. The OH of the phenol should be at para position of linkage
Pthalein dye test diff colours for diff phenolic grps
phenol: pink o-cresol: red m-cresol: bluish purple p-cresol: no color catechol: blue resorcinol: green
diazo + B napthol
scarlet red dye(sparingly soluble in water)
B napthol is two benzene rings and one of them has OH on metaposition of attachment)
The diazo gets coupled on the OH ring between the OH and the attachment(of the other ring)
Solubility in HCl is a test for
amines in general
Preparation of Schiff’s reagent
decolorizing p-rosaniline hydrochloride dye by adding soium sulphite or by passing SO2 gas
Claisen Schmidt reaction of aromatic aldehyde with aldehyde/ketone(having alpha hydrogen)
acetone + 2benzaldehyde —-> dibenzalacetone
Replace O of C=O of benzaldehyde with two alpha hydrogens of acetone
And then do the same for other side
nitrating mixture
HNO3 + H2SO4 ——> NO2+ + H3O+ + HSO4-
aniline yellow
diazo salt + aniline
happens at para position
Molisch’s test reagent
conc H2SO4 + alcoholic solution of 1 naphthol
violet color appears
Barfoed test reagent
Cupric acetate in acetic acid solution
brick red ppt of Cu2O observed
