Why does the reaction of 2-bromo-2-methylbutane and ethanolic solution of silver nitrate interfered by the presence of water in the sodium iodide in acetone reagent?
(Substitution chemistry)
Thanks for any help!!
2-bromo-2-methybutane reacts with AgNO3 to form AgBr precipitate.
If NaI is present, it also reacts reacts with AgNO3 to form AgI precipitate.
When you observe the precipitate, how would you know what caused it?
In the reaction of 2-bromo-2-methylbutane (also known as t-butyl bromide) with an ethanolic solution of silver nitrate, a substitution reaction takes place. The bromine atom in the t-butyl bromide is replaced by the nitrate ion from the silver nitrate, resulting in the formation of t-butyl nitrate.
However, the presence of water in the sodium iodide in acetone reagent interferes with this reaction. In this reagent, sodium iodide (NaI) is dissolved in acetone, forming a solution. Normally, NaI in acetone acts as a nucleophile, attacking the carbon atom bonded to the bromine in the t-butyl bromide. This leads to a substitution reaction and the formation of t-butyl iodide.
But when water is present in the NaI in acetone reagent, it can react with the silver nitrate to form silver chloride (AgCl) and nitric acid (HNO3). The formation of AgCl is a precipitation reaction, resulting in the formation of a white insoluble precipitate. This precipitation reaction consumes the Ag+ ions, which are the active species responsible for the substitution reaction.
As a result, the presence of water in the NaI in acetone reagent leads to a decrease in the concentration of Ag+ ions available for the substitution reaction, inhibiting or interfering with the reaction between t-butyl bromide and the ethanolic solution of silver nitrate. This interference can significantly reduce the yield of the desired t-butyl iodide product in this substitution reaction.