Compound 'A' with molecular formula `C_(4)H_(9)Br` is treated with aq.KOH solution. The rate of this reaction depends upon the concentration of the compound 'A' only. When another optically active isomer 'B' of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.
(i) Write down the structural formula of both compounds 'A' and 'B'.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

Explanation

The rate of reaction of compound `A(Me:C_(4)H_(9)Br)` with aq. KOH depends upon the concentration of the compound 'A' only. So the reaction occurs by `S_(N)1)` pathway. This shows that the compound 'A' is t-butyl bromide. `[(CH_3)_(3)CBr]`. Now, compound 'B' is an optically active isomer of compound 'A' . so compound B must be 2-borombutane. The rate of reaction of compound B with aq.KOH depends on the concentration of both substance and nucleophile, so the reaction proceeds by `S_(N)2` mechanism, causing inversion of configuration.