An alcohol [A] with molecules formula `(C_(4)H_(10)O)` o oxidation with aciddified potassium dichromate gives acid [B] `(C_(4)H_(8)O_(2))`. Compound [A] when dehydrated with conc. `H_(2)SO_(4)` at `443 K` gives compound [C]. Treatment of [C] with aqueous `H_(2)SO_(4)` gives compound [D] `(C_(4)H_(10)O)` which is an isomer of [A]. compound [D] is resistant to oxidation but compound [A]can be easily oxidised. Identify [A], [B], [C] and [D]. Name the type of isomerism exhibited by [A] and [D].

Explanation

To solve the question, we need to identify the compounds A, B, C, and D based on the given information and their molecular formulas. Let's break it down step by step:

### Step 1: Identify Compound A
- **Given**: Compound A has the molecular formula \( C_4H_{10}O \).
- **Analysis**: This formula corresponds to a secondary or tertiary alcohol since it can be oxidized to form a carboxylic acid.
- **Possible Structure**: The most likely candidate for compound A is **2-methylpropan-1-ol** (or **tert-butanol**), which is \( CH_3C(OH)(CH_3)CH_2 \).

### Step 2: Identify Compound B
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