An alkyl bromide A has molecular formula `C_(8)H_(17)Br` and four different structures can be drawn for it, all optically active. A on refluxing with ethanolic `KOH` solution yields only one elirnination product `B(C_(8)H_(16))` which it still enantiomeric. B on treatment with `H_(2)//Pt` yields `C(C_(8)H_(18))` which does not rotate the plane polarized light, B on ozonolysis followed by work-up with `H_(2)O_(2)` yields `D(C_(7)H_(14)O)` as one product which it still resolvable. Deduce structures of A to D.

Answer

To solve the problem step by step, we will deduce the structures of compounds A, B, C, and D based on the information provided.

### Step 1: Identify the structure of A
The molecular formula of A is \( C_8H_{17}Br \). Given that A has four different structures that are all optically active, we can deduce that A must have at least one chiral carbon.

Let's consider the following structure for A:
1. **Structure of A**:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]
This structure has a chiral center at the carbon atom bonded to Br.

### Step 2: Determine the elimination product B
When A is refluxed with ethanolic KOH, it undergoes an elimination reaction to form B, which has the formula \( C_8H_{16} \). The elimination reaction typically results in the formation of a double bond.

2. **Structure of B**:
\[ \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]
This structure of B is still optically active and can exist as enantiomers.

### Step 3: Determine the product C from B
When B is treated with \( H_2 \) in the presence of Pt, it undergoes hydrogenation to yield C, which has the formula \( C_8H_{18} \). The product C is optically inactive, indicating that it has a symmetrical structure.

3. **Structure of C**:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]
This is n-octane, which is symmetrical and does not rotate plane-polarized light.

### Step 4: Determine the product D from ozonolysis of B
When B undergoes ozonolysis followed by work-up with \( H_2O_2 \), it yields D, which has the formula \( C_7H_{14}O \). The ozonolysis typically cleaves the double bond and forms carbonyl compounds.

4. **Structure of D**:
\[ \text{CH}_3-\text{CH}_2-\text{C}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]
This product is still resolvable, indicating it has at least one chiral center.

### Summary of Structures
- **A**: \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_2-\text{CH}_3 \)
- **B**: \( \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \)
- **C**: \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \) (n-octane)
- **D**: \( \text{CH}_3-\text{CH}_2-\text{C}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{CH}_3 \)