Compound 'A' with molecular formula `C_(4)H_(9)`Br is treated with aq. KOH solution. The rate of this reaction depends upon the the concentration of the compounds '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

To solve the problem, we will analyze the information given about the compounds A and B, derive their structures, and determine which compound undergoes inversion of configuration upon reaction with aqueous KOH.

### Step 1: Identify Compound A
Given the molecular formula \( C_4H_9Br \), we need to identify a compound that is likely to undergo an SN1 reaction, as the rate of reaction depends only on the concentration of compound A.

- **Structure of Compound A**:
A tertiary alkyl bromide is likely to fit this description. The structure can be:

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