in alkynes reactions, what is the function of Baeyer test and Bromine test and their expected result?

Answer 1

to test for the presence of carbon-carbon triple bonds. The result of the Baeyer test is a 4 hydroxy quattrol. The purple KMn04 should turn colorless, and a brown solid suspension of Mn02 will remain in the solution.

bromine will do the same thing, it will add across the triple bond changing it into an alkene, and with the addition of more bromine it will reduce it down to a alkane. If its a positive test, the color of the bromine will disapear and there will be NO gas evolution.