Procedure No. 1 Reaction Dissolve 50 mg of the starting material in 3 mL of acet

Question

Procedure No. 1 Reaction Dissolve 50 mg of the starting material in 3 mL of acetone using a 10 mL round-bottom flask. Use 3 to 5 mL of this sample as a control for thin layer chromatography. Add to the flask 0.12 mL of glacial acetic acid, and start the reaction by adding 0.4 mL of a 5.25% sodium hypochlorite solution (commercial bleach). After 5 min analyze the reaction progress by assaying the mixture for the presence of the starting material by TLC using a mobile phase of 30% acetone and 70% hexane on fluorescent coated silica plates visualized using UV lights. If the reaction is not finished (as determined by TLC), 0.4 mL increments of bleach are added until the reaction is judged to be complete. Once the conversion is complete and the reaction sample shows no more starting material by TLC analysis, proceed with an extraction of the product. Extract this crude mixture two times, each with 2 mL hexane. Then wash the hexane extract with 1 mL of 5% sodium bicarbonate solution to eliminate any acetic acid and then with 2 mL of water. Dry the hexane extract with an appropriate anhydrous drying agent for a few minutes, and transfer the hexane extract to a previously weighed round bottomed flask. Evaporate the hexane on the rotary evaporator to isolate the crude product. If necessary, the product may be recrystallized from hexane.

Explanation / Answer

The reaction between glacial acetic acid and sodium hypochlorite is acid-base reaction as follows:

NaOCl +CH3COOH -----> HOCl +CH3COO- Na+

Starting material (50 mg)+ HOCl ------> product

The HOCl is possible source of positively charged Cl+ ion. but positive chloride ion has not been found descrete in solution.So, there may be possibility to form Cl- by electron transfer.

The negatively charged chlorine (Cl-) proceed the reaction as oxidation reaction(loss of electron) .

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