When `CH_(3)CH_(2)CHCl_(2)` is treated with `"NaNH"_(2)` the product formed is:

Options

A. `CH_(3)` - CH = `CH_(2)`
B. `CH_(3) - C -=CH`
C.
D.

Answer

To solve the problem of what product is formed when `CH₃CH₂CHCl₂` (1,1-Dichloropropane) is treated with sodium amide (`NaNH₂`), we can follow these steps:

### Step-by-step Solution:

1. **Identify the Reactant**: The reactant is 1,1-Dichloropropane, which has the structure `CH₃CH₂CHCl₂`. This means that there are two chlorine atoms attached to the same carbon atom.

2. **Understand the Reagent**: Sodium amide (`NaNH₂`) is a strong base. Its primary role in this reaction is to abstract protons (H⁺) from the molecule.

3. **Identify Acidic Protons**: In 1,1-Dichloropropane, the carbon atom bonded to the chlorine atoms (the second carbon in the chain) has acidic protons. These protons can be abstracted by the strong base.

4. **Proton Abstraction**: When `NaNH₂` abstracts a proton from the carbon adjacent to the carbon with chlorine, it forms a carbanion. This carbanion is stabilized by the presence of the electronegative chlorine atom.

5. **Elimination of HCl**: The abstraction of the proton leads to the elimination of a chlorine atom as HCl. This results in the formation of a double bond (alkene) in the molecule.

6. **Further Dehydrohalogenation**: The alkene that is formed can undergo further reaction with `NaNH₂`, leading to another elimination reaction where another HCl is removed. This results in the formation of an alkyne.

7. **Final Product**: The final product after the complete reaction is 1-Propyn-1, which has the structure `CH₃C≡C`.

### Conclusion:
The product formed when `CH₃CH₂CHCl₂` is treated with `NaNH₂` is 1-Propyn-1.