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 equiv CH`
C.
D.

Answer

To solve the problem of what product is formed when `CH3CH2CHCl2` is treated with `NaNH2`, we can follow these steps:

### Step 1: Identify the Reactants
The reactant is `CH3CH2CHCl2`, which is 1,1-dichloropropane. The reagent is sodium amide (`NaNH2`), a strong base.

**Hint:** Recognize the structure of the reactant and the nature of the reagent.

### Step 2: Understand the Reaction Mechanism
`NaNH2` is a strong base that can deprotonate hydrogen atoms and eliminate halides. In this case, it will first remove one of the chlorine atoms and a hydrogen atom from the carbon chain.

**Hint:** Remember that strong bases can lead to elimination reactions, resulting in the formation of alkenes or alkynes.

### Step 3: First Elimination
When `NaNH2` reacts with `CH3CH2CHCl2`, it can remove one `Cl` atom and one `H` atom from the adjacent carbon, leading to the formation of a double bond. The first step results in the formation of `CH3CH=CHCl` (propene with a chlorine substituent).

**Hint:** Look for the formation of a double bond after the elimination of HCl.

### Step 4: Second Elimination
The product `CH3CH=CHCl` can further react with `NaNH2`. The strong base can again remove the remaining `Cl` and a hydrogen atom adjacent to it, leading to the formation of a triple bond. This results in the final product `CH3C≡CH` (propyne).

**Hint:** Consider the possibility of multiple elimination reactions when a strong base is involved.

### Step 5: Final Product
The final product after both eliminations is `CH3C≡CH`, which is propyne.

**Hint:** Ensure that you account for all the halides and hydrogens that are removed during the reaction.

### Conclusion
When `CH3CH2CHCl2` is treated with `NaNH2`, the final product formed is `CH3C≡CH` (propyne).

**Final Answer:** The product formed is `CH3C≡CH` (option B).