Question

Why is it important to completely dissolve both the alkene and alkyne before adding the pyridiumium tribromide to the reaction vial. Why is this important? Hint: the alkene and alkyne do not have the same solubility in acetic acid.

this is the experiment,

In the fume hood, measure 2.0 mL of glacial acetic acid into your 10 mL microscale reaction vial. Add trans-stilbene (135 mg, 0.75 mmol) and diphenylacetlyene (135 mg, 0.75 mmol). Add a stir bar and place the vial in the heating block set at 70 °C. Once all of the solid has dissolved, carefully add pyridinium tribromide (240 mg, 0.75 mmol) to the reaction vial. Continue heating the reaction mixture for 15 minutes. In your notebook, note any changes that occur once the pyridinium tribromide is added. While the reaction is running, use your small reaction vial to heat about 1 mL of water in to about 70 °C. When the 15 minute reaction time is complete, add about 0.5 mL of your warmed water to the reaction mixture. Allow the reaction mixture to stir for 1-2 minutes. Lift the reaction vial from the heating block, allow it to cool to room temperature, and then cool it further in an ice water bath for about 5 minutes. Collect the resulting crystals by vacuum filtration using a Hirsch funnel and wash them with about 1mL of cold water. Dry the collected solid by pulling air through it with the vacuum for about 5 min. Place the solid in a tared, labeled sample vial. Determine the mass of the sample and submit it to your instructor for GC-MS analysis.

Answer 1

The other reagent in this reaction is Bromination agent Pyridinium tribromide. The Br2 Bromine is generated in-situ. And hence all the reactants must be in same phase to have maximum reactivity, and optimum yield.

As the Br2 is in solution, we need to have alkene and alkyne in the solution form as well. The solubility of these are different in acetic acid and hence after addition the mixture is heated to make everything solubilized in the solution and give clear solution.