The isomeric deuterated bromide with molecular formula `C_(4)H_8DBr` having two chiral carbon atoms is

Options

A. `2` - Bromo `-1` - deuterobutane`
B. 2-Bromo - `2` - douterobutane
C. `2` - Bromo `-3` - deuterobutane
D. `2` - Bromo `-1` - deuteron - `2` - methylpropane

Answer

To solve the problem of identifying the isomeric deuterated bromide with the molecular formula \( C_4H_8DBr \) that has two chiral carbon atoms, we can follow these steps:

### Step 1: Analyze the Molecular Formula
The molecular formula given is \( C_4H_8DBr \). This indicates that we have:
- 4 carbon atoms (C)
- 8 hydrogen atoms (H)
- 1 deuterium atom (D)
- 1 bromine atom (Br)

### Step 2: Determine the Structure
Given that we need to have two chiral centers, we can start by constructing a potential structure. A common structure that fits this formula and has two chiral centers is:
\[ CH_3-CHBr-CHD-CH_3 \]

### Step 3: Identify Chiral Centers
In the structure \( CH_3-CHBr-CHD-CH_3 \):
- The second carbon (attached to Br) is chiral because it has four different groups: \( CH_3 \), \( Br \), \( CHD \), and \( H \).
- The third carbon (attached to D) is also chiral because it has four different groups: \( CH_3 \), \( H \), \( CHBr \), and \( D \).

### Step 4: Confirm the Isomer
Now, we need to confirm that this structure is indeed the isomer we are looking for. The compound can be named as:
- 2-bromo-3-deuterobutane

### Step 5: Final Answer
Thus, the isomeric deuterated bromide with the molecular formula \( C_4H_8DBr \) having two chiral carbon atoms is:
**2-bromo-3-deuterobutane**