In the reaction
`CH_(3) C -= C^(-) - Na^(+) +(CH_(3))_(2) CH - Cl to ?`
the product formed is .

Options

A. propene
B. propyne
C. propyne and propene
D. 4 - methylpentyne - 2

Explanation

To solve the reaction given in the question, we need to analyze the reactants and the type of reaction taking place.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are:
- An alkynyl anion: `CH3C≡C^(-)Na^(+)` (which is sodium acetylide)
- An alkyl halide: `(CH3)2CH-Cl` (which is 2-chloro-2-methylpropane)

2. **Recognize the Reaction Type**:
This reaction is a nucleophilic substitution reaction (S_N2 mechanism), where the nucleophile (the alkynyl anion) will attack the carbon atom bonded to the chlorine in the alkyl halide, leading to the formation of a new carbon-carbon bond.

3. **Mechanism of the Reaction**:
- The alkynyl anion (`CH3C≡C^(-)`) will attack the carbon atom of the alkyl halide (`(CH3)2CH-Cl`), displacing the chloride ion (Cl^-) and forming a new compound.
- The carbon atom in the alkyl halide that is bonded to chlorine is a tertiary carbon, which is favorable for the S_N2 reaction due to steric hindrance.

4. **Product Formation**:
- After the nucleophilic attack, the product formed will be `CH3C≡C-CH(CH3)2`, which is 4-methyl-2-pentyne.
- The triple bond is located between the second and third carbon atoms in the chain.

5. **Naming the Product**:
- The longest carbon chain containing the triple bond has five carbon atoms, making it a pentyne.
- The methyl group is on the fourth carbon when numbering from the end closest to the triple bond.
- Therefore, the IUPAC name of the product is **4-methyl-2-pentyne**.

### Final Answer:
The product formed in the reaction is **4-methyl-2-pentyne**.