When t-butyl bromide is treated with CH_(3)O^(-) in a mixture of CH_(3)OHandH_(2)O, the rate of formation of t-butyl alcohol and t-butyl methyl ether doesn't change appreciably as the concentration of CH_(3)O^(-) ion is increased. However, increasing the concentration of CH_(3)O caused a marked increase in the rate at which t-butyl bromide disappear from the mxiture. Why?

When t-butyl bromide is treated with CH_(3)O^(-) in a mixture of CH_(3

1.	When t-butyl bromide is treated with CH_(3)O^(-) in a mixture of CH_(3)OHandH_(2)O, the rate of formation of t-butyl alcohol and t-butyl methyl ether doesn't change appreciably as the concentration of CH_(3)O^(-) ion is increased. However, increasing the concentration of CH_(3)O caused a marked increase in the rate at which t-butyl bromide disappear from the mxiture. Why?
Answer» Solution :t-Butyl alcohol and t-butyl METHYL ether are FORMED through `S_(N)1` MECHANISM and the rate of the reactions in independent of the concentration of `CH_(3)O^(-)`. This was the only reaction that causes t-butyl bromide to disapear. But if we increase the concentration of `CH_(3)O^(-)`,t-butyl bromide STARTS disappearing through E2 mechanism. This reaction is DEPENDENT on the concectration of `CH_(3)O^(-)`. Hence as we increase teh concentration of `CH_(3)O^(-)` the rate of disappearance of t-butyl bromide increase the concentration of `CH_(3)O_(-)` the rate of disappearance of t-butyl bromide increases.