In the reaction
`CH_(3)C equiv C^(-)Na^(+) + (CH_(3))_(2) `CHCl `rarr` Product the product fomed is :

Options

A. 4-Methyl-2-pentyne
B. Propyne
C. Propyne and propene
D. None of these

Answer

To solve the problem, we need to analyze the reaction between the sodium salt of an alkyne and 2-chloropropane. Here’s a step-by-step breakdown:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Sodium salt of an alkyne: \( CH_3C \equiv C^{-}Na^{+} \)
- 2-Chloropropane: \( (CH_3)_2CHCl \)

### Step 2: Understand the Structure of the Reactants
1. **Sodium Salt of Alkyne**:
- The structure can be represented as:
\[ CH_3-C \equiv C^{-} \quad (with \, Na^{+}) \]
- This has a nucleophilic carbon atom (the one with the negative charge).

2. **2-Chloropropane**:
- The structure is:
\[ CH_3-CH(Cl)-CH_3 \]
- The carbon atom attached to chlorine is electrophilic due to the electronegativity of chlorine.

### Step 3: Reaction Mechanism
- The nucleophilic carbon from the sodium salt of the alkyne will attack the electrophilic carbon in 2-chloropropane.
- This results in the displacement of the chlorine atom (a nucleophilic substitution reaction).

### Step 4: Formation of the Product
- The product formed will be an extended alkyne:
\[ CH_3-C \equiv C-CH(CH_3)-CH_3 \]
- The sodium chloride (NaCl) is released as a byproduct.

### Step 5: Naming the Product
1. **Identify the Longest Carbon Chain**: The longest chain contains 5 carbon atoms.
2. **Number the Chain**: Start numbering from the end nearest to the alkyne functional group:
- The numbering will be: 1 (alkyne) - 2 - 3 - 4 - 5.
3. **Identify Substituents**: There is a methyl group at position 4.
4. **IUPAC Name**: The name of the compound will be:
- 4-methyl-pent-2-yne.

### Final Answer
The product formed from the reaction is **4-methyl-pent-2-yne**.

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