An alkyl halide (A) of formula `C_(6)H_(11)Cl` on treatement with potassium tertiary butoxie gives two isomeric alkenes (B) and (C) `C_(6)H_(10)`. Both alkene on hydrogenation give methycyclopentane. Predict the structure of (A), (B) and (C).

Answer

To solve the problem, we need to identify the structures of the alkyl halide (A) and the two isomeric alkenes (B) and (C) based on the given information. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the Degree of Unsaturation
The formula for the degree of unsaturation (DU) is given by:

\[ \text{DU} = \frac{(2C + 2 - H - X)}{2} \]

Where:
- \(C\) = number of carbon atoms
- \(H\) = number of hydrogen atoms
- \(X\) = number of halogen atoms

For compound A, \(C_6H_{11}Cl\):
- \(C = 6\)
- \(H = 11\)
- \(X = 1\) (for Cl)

Substituting these values into the formula:

\[ \text{DU} = \frac{(2 \times 6 + 2 - 11 - 1)}{2} = \frac{(12 + 2 - 11 - 1)}{2} = \frac{2}{2} = 1 \]

### Step 2: Interpret the Degree of Unsaturation
A degree of unsaturation of 1 indicates either a double bond or a ring structure. Given that the final product after hydrogenation is methylcyclopentane, it suggests that the structure likely contains a ring.

### Step 3: Propose Possible Structures for A
Since we have a 6-carbon alkyl halide, we can propose several structures for A. The possible structures for A could include:

1. **1-Chloro-2-methylpentane**
2. **2-Chloro-2-methylpentane**
3. **3-Chloro-2-methylpentane**
4. **1-Chloro-3-methylpentane**

### Step 4: Elimination Reaction with Potassium Tert-Butoxide
When treated with potassium tert-butoxide, an elimination reaction occurs, leading to the formation of alkenes. We need to consider the elimination of HCl to form alkenes (B and C).

1. From **1-Chloro-2-methylpentane**, we can remove H from C-2 or C-3 leading to:
- **Alkene B**: 2-methyl-1-pentene
- **Alkene C**: 3-methyl-1-pentene

2. From **2-Chloro-2-methylpentane**, we can only remove H from C-1 or C-3 leading to:
- **Alkene B**: 2-methyl-1-pentene
- **Alkene C**: 2-methyl-2-pentene

3. From **3-Chloro-2-methylpentane**, we can remove H from C-2 or C-4 leading to:
- **Alkene B**: 3-methyl-1-pentene
- **Alkene C**: 2-methyl-2-pentene

### Step 5: Hydrogenation of Alkenes
Both alkenes B and C yield methylcyclopentane upon hydrogenation. This indicates that both alkenes must be capable of forming the same cyclopentane structure upon reduction.

### Final Structures
- **Structure of A**: 1-Chloro-2-methylpentane
- **Structure of B**: 2-methyl-1-pentene
- **Structure of C**: 3-methyl-1-pentene

### Summary of Structures
- **A**: 1-Chloro-2-methylpentane
- **B**: 2-methyl-1-pentene
- **C**: 3-methyl-1-pentene