| Demonstration Experiment on Video Objectives: Nucleophilic Carbonyl alpha-Substitution, Test for the alpha-Methyl Carbonyl Group Peter Keusch |
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German version
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.
Hazards and safety precautions:
Safety glasses, nitrile gloves and effective ventilation required. The 0.2 M iodine solution should be prepared inside a fumehood! 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
|  60 mL of ethanol are added to the iodine solution in conical measure 1. Afterwards 50 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. 
 Video clip (Download RealPlayer .rm file)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 CH3 - CO group. 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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