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What is the technique for iodoform test for aromatic compounds?????? Plz. give a detailed illustration!!!!!!!
Dear student,
The iodoform test or reaction is a chemical reaction where iodoform (CHI3) is produced by the multiple halogenation of a methyl ketone (a molecule containing the R-CO-CH3 group) in the presence of a base . R may be H, alkyl or aryl. The reaction can be used to produce CHCl3, CHBr3 or CHI3. In analytical chemistry, this reaction was traditionally used to determine the presence of a methyl ketone, or a secondary alcohol oxidizable to a methyl ketone through the the iodoform test. Nowadays, spectroscopic techniques such as NMR and infrared spectroscopy are preferred because they require small samples, may be non-destructive (for NMR) and are easy and quick to perform. Formerly, it was used to produce iodoform and bromoform and even chloroform industrially. In organic chemistry, this reaction may be used to convert a terminal methyl ketone into the appropriate carboxylic acid.
dear aswin , the main thng which should be kept in mind is that the compound should have 3 alfa acidic hydrogens(presence of methyl group) ...
this is the only condition ...
example => acetone, acetophenone , acetaldehyde , ....
case1) if there are only two alfa hydoogens & Iodo group present at alfa pos then this will give iodoform rxn...
case 2) if there are only 1 alfa hydorgen & 2 Iodo groups then also this will give this test
case3) if there are no acidic hydrogens but 3 Iodo groups present then also this will give Iodoform test ...
there is no particular change for aeromatic or aliphatic , the need is three acidic hydorgens only....
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