`C_(5)H_(11)Br+ NaCN rarr A overset(H_(3)O^(+))(rarr) B + "NaOH" overset("CaO")(rarr) `C
C' has the formula `C_(5)H_(12)` which can give four structural isomeric monochloro derivative . What is the structure of `C_(5)H_(11) Br` ?

Options

A. `CH_(3)- underset(CH_(3))underset(|)(CH) - CH_(2) CH_2- `Br
B. `CH_(3) - CH_(2) - overset(Br)overset(|)(CH) - CH_(2) - CH_(5)`
C. `CH_(5) - CH_(2) - CH_(2) - underset(Br)underset(|)(CH) - CH_(3)`
D. `CH_(2) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C ) - CH_(2) - `Br

Answer

To solve the question regarding the structure of \( C_5H_{11}Br \), we will follow a systematic approach based on the information provided.

### Step 1: Determine the Degree of Unsaturation (DB)
The degree of unsaturation (DB) can be calculated using the formula:
\[ DB = \frac{C + 1 - H - N + X}{2} \]
Where:
- \( C \) = number of carbon atoms
- \( H \) = number of hydrogen atoms
- \( N \) = number of nitrogen atoms
- \( X \) = number of halogen atoms

For \( C_5H_{11}Br \):
- \( C = 5 \)
- \( H = 11 \)
- \( N = 0 \)
- \( X = 1 \) (since we have one bromine atom)

Substituting the values:
\[ DB = \frac{5 + 1 - 11 - 0 + 1}{2} = \frac{-4}{2} = 0 \]
This indicates that there are no pi bonds or rings in the structure, suggesting it is either a straight-chain or branched alkyl halide.

**Hint:** The degree of unsaturation helps identify if there are rings or double bonds in the molecule.

### Step 2: Reaction with Sodium Cyanide (NaCN)
The reaction of \( C_5H_{11}Br \) with NaCN will replace the bromine atom with a cyanide group:
\[ C_5H_{11}Br + NaCN \rightarrow C_5H_{11}CN \]
This results in the formation of a nitrile compound.

**Hint:** The nucleophilic substitution reaction replaces the halogen with a cyanide group.

### Step 3: Hydrolysis of the Nitrile
Hydrolysis of the nitrile \( C_5H_{11}CN \) in the presence of \( H_3O^+ \) will convert the nitrile into a carboxylic acid:
\[ C_5H_{11}CN + H_3O^+ \rightarrow C_5H_{11}COOH \]
This step involves breaking the carbon-nitrogen triple bond and forming a carboxylic acid.

**Hint:** Hydrolysis of nitriles typically yields carboxylic acids.

### Step 4: Decarboxylation
When the carboxylic acid \( C_5H_{11}COOH \) is treated with NaOH and heated with CaO, decarboxylation occurs, leading to the formation of an alkene:
\[ C_5H_{11}COOH \xrightarrow{NaOH, \, CaO} C_5H_{12} + CO_2 \]
This results in the formation of an alkene with the formula \( C_5H_{12} \).

**Hint:** Decarboxylation reactions remove carbon dioxide and often yield alkenes.

### Step 5: Identify the Structure of \( C_5H_{12} \)
The final compound \( C_5H_{12} \) can form four structural isomeric monochloro derivatives. This indicates that the structure must have sufficient branching to allow for different positions of the chlorine atom.

The possible structure of \( C_5H_{11}Br \) that can lead to four isomeric monochloro derivatives is:
- **2-Bromopentane**: This structure allows for chlorination at four different carbon positions.

**Hint:** The ability to form multiple isomers indicates that the compound has a branched structure.

### Conclusion
The structure of \( C_5H_{11}Br \) is likely **2-Bromopentane**, which can yield four structural isomers upon chlorination.