Objectives: Nucleophilic Carbonyl alpha-Substitution,
Test for the alpha-Methyl Carbonyl Group
Chemicals:
potassium iodide
iodine
ethanol 95%
2-propanol
acetone
2 N NaOH
Preparation of a 0.2 M iodine solution:
Using an Erlenmeyer flask, 37.35 g of potassium iodide are dissolved in 375 mL of dist. water. After the addition of 19.05 g of iodine the mixture must be stirred until it is homogeneous.
Apparatus and glass wares:
hot plate
2 thermometers
3 conical measures, graduated, 500 mL
3 glass stirring rods
Erlenmeyer flask 500 mL
3 beakers 200 mL
1 beaker 100 mL
2 pipettes 5 mL, graduated in 0.1 mL
1 pipette bulb
2 snap-cap vials 20 mL
2 measuring cylinders 200 mL
Experimental procedure:
conical measure 1: 200 ml of 0.2 M iodine solution, warmed up to 40°C
conical measure 2: 200 ml of 0.1 M iodine solution
conical measure 3: 200 ml of 0.1 M iodine solution
Iodoform test on ethanol:
60 mL of ethanol are added to the iodine solution in conical measure 1. Afterwards 150 mL of 2 N NaOH (warmed up to 40°C) are added while stirring.
Iodoform test on 2-propanol:
The iodine solution in conical measure 2 is mixed with 3.8 mL of 2-propanol and 150 mL of 2 N NaOH (room temperature).
Iodoform probe on acetone:
3.7 mL of acetone and 150 mL of 2 N NaOH (room temperature) are added to the iodine solution in conical measure 3.
Results:
A yellow, crystalline precipitate is formed in each of the three conical measures.
Discussion:
The iodoform reaction is characteristic for methylketones as well as for alcohols (e.g. ethanol, 2-propanol), that can be oxidized to a methyl carbonyl compounds. The iodoform test is a test for the existence of the CH3-CO group in a molecule. The group to which the CH3-CO group is attached can be aryl, alkyl and hydrogen.
Both ethanol and 2-propanol are oxidized by iodine to give ethanale or acetone. (1).
When a-methyl carbonyl compounds react with iodine in the presence of a base, the hydrogen atoms on the carbon adjacent to the carbonyl group (a hydrogens) are subsituted by iodine to form tri iodo methyl carbonyl compounds which react with OH - to produce iodoform and carboxylic acid (2):
Reaction mechanism:
The hydrogen atoms on the methyl group are slightly acidic and can be removed with hydroxide. The carbanion formed then react with iodine molecules to give a iodide ion and a monoiodonated methyl carbonyl derivate. Introduction of the first iodine atom (owing to its electronegativity) makes the remaining hydrogens of the methyl group more acidic. Hence a base-catalized iodination of a monohalogenated methyl carbonyl derivate occurs at the carbon that is already substituted. Finally a tri iodo methyl carbonyl derivate is formed.
The next step is a nucleophilic attack by hydroxide on the carbonyl carbon atom. A carbon-carbon bond cleavage occurs and a triiodomethanide ion departs. The triiodomethanide ion is unusually stable. Its negative charge is dispersed by the three negative iodine atoms. In the last step a proton transfer takes place between carboxylic acid and triiodomethanide ion to form ultimately carboxylate ion and iodoform.
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