When `CH_3CH_2CHCl_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 determine the product 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. It contains two chlorine atoms attached to the same carbon atom (the second carbon in the chain).

### Step 2: Understand the role of NaNH2
Sodium amide (`NaNH2`) is a strong base. In the presence of a strong base, it can facilitate the elimination of halogens (in this case, chlorine) from the alkyl halide.

### Step 3: Mechanism of reaction
The reaction involves a double elimination process:
1. The strong base `NaNH2` abstracts a proton (H) from the carbon adjacent to the carbon with the halogen (Cl).
2. This leads to the formation of a double bond as the chlorine atoms are eliminated as HCl.

### Step 4: Perform the elimination
In `CH3CH2CHCl2`, the two chlorine atoms can be removed along with two hydrogen atoms from the adjacent carbon:
- The first elimination removes one Cl and one H, forming a double bond.
- The second elimination removes the second Cl and another H, resulting in a triple bond.

### Step 5: Determine the product
After the elimination of two HCl molecules, we are left with a carbon chain that has a triple bond between the first and second carbon atoms. The resulting product is propyne (`CH3C≡C`).

### Conclusion
The product formed when `CH3CH2CHCl2` is treated with `NaNH2` is **propyne** (`C3H4` or `CH3C≡CH`).

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