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 step by step, we will analyze the information given about the compounds 'A' and 'B', their structures, and the mechanisms involved in their reactions with aqueous KOH.

### Step 1: Identify Compound 'A'
Given the molecular formula \(C_4H_9Br\), we need to determine the structure of compound 'A'. Since the reaction with aqueous KOH depends only on the concentration of compound 'A', it indicates that the reaction proceeds via the SN1 mechanism. This suggests that compound 'A' is a tertiary alkyl halide, as tertiary carbocations are more stable.

The structure of compound 'A' can be represented as:
- **Compound A**: 2-bromobutane (Tertiary bromide)

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